Solve x2 9

Solve x2 9 DEFAULT

Contents: This page corresponds to §2.4 (p. 200) of the text.

Suggested Problems from Text:

p. 212 #7, 8, 11, 15, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, 60, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97

Quadratic Equations

Equations Involving Radicals

Polynomial Equations of Higher Degree

Equations Involving Fractional Expressions or Absolute Values


Quadratic Equations

A quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0.

Factoring

This approach to solving equations is based on the fact that if the product of two quantities is zero, then at least one of the quantities must be zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, see the review section P.3 (p.26) of the text.

Example 1.

2x2 - 5x - 12 = 0.

(2x + 3)(x - 4) = 0.

2x + 3 = 0 or x - 4 = 0.

x = -3/2, or x = 4.

Square Root Principle

If x2 = k, then x = ± sqrt(k).

Example 2.

x2 - 9 = 0.

x2 = 9.

x = 3, or x = -3.


Example 3.


Example 4.

x2 + 7 = 0.

x2 = -7.

x = ± .

Note that = = , so the solutions are

x = ± , two complex numbers.

Completing the Square

The idea behind completing the square is to rewrite the equation in a form that allows us to apply the square root principle.

Example 5.

x2 +6x - 1 = 0.

x2 +6x = 1.

x2 +6x + 9 = 1 + 9.

The 9 added to both sides came from squaring half the coefficient of x, (6/2)2 = 9. The reason for choosing this value is that now the left hand side of the equation is the square of a binomial (two term polynomial). That is why this procedure is called completing the square. [ The interested reader can see that this is true by considering (x + a)2 = x2 + 2ax + a2. To get "a" one need only divide the x-coefficient by 2. Thus, to complete the square for x2 + 2ax, one has to add a2.]

(x + 3)2 = 10.

Now we may apply the square root principle and then solve for x.

x = -3 ± sqrt(10).


Example 6.

2x2 + 6x - 5 = 0.

2x2 + 6x = 5.

The method of completing the square demonstrated in the previous example only works if the leading coefficient (coefficient of x2) is 1. In this example the leading coefficient is 2, but we can change that by dividing both sides of the equation by 2.

x2 + 3x = 5/2.

Now that the leading coefficient is 1, we take the coefficient of x, which is now 3, divide it by 2 and square, (3/2)2 = 9/4. This is the constant that we add to both sides to complete the square.

x2 + 3x + 9/4 = 5/2 + 9/4.

The left hand side is the square of (x + 3/2). [ Verify this!]

(x + 3/2)2 = 19/4.

Now we use the square root principle and solve for x.

x + 3/2 = ± sqrt(19/4) = ± sqrt(19)/2.

x = -3/2 ± sqrt(19)/2 = (-3 ± sqrt(19))/2

So far we have discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right form to apply the square root principle may be rearranged and solved by factoring as we see in the next example.

Example 7.

x2 = 16.

x2 - 16 = 0.

(x + 4)(x - 4) = 0.

x = -4, or x = 4.

In some cases the equation can be solved by factoring, but the factorization is not obvious.

The method of completing the square will always work, even if the solutions are complex numbers, in which case we will take the square root of a negative number. Furthermore, the steps necessary to complete the square are always the same, so they can be applied to the general quadratic equation

ax2 + bx + c = 0.

The result of completing the square on this general equation is a formula for the solutions of the equation called the Quadratic Formula.

Quadratic Formula

The solutions for the equation ax2 + bx + c = 0 are

We are saying that completing the square always works, and we have completed the square in the general case, where we have a,b, and c instead of numbers. So, to find the solutions for any quadratic equation, we write it in the standard form to find the values of a, b, and c, then substitute these values into the Quadratic Formula.

One consequence is that you never have to complete the square to find the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so you still need to know how to do it!

Examples using the Quadratic Formula:

Example 8.

2x2 + 6x - 5 = 0.

In this case, a = 2, b = 6, c = -5. Substituting these values in the Quadratic Formula yields

Notice that we solved this equation earlier by completing the square.

Note: There are two real solutions. In terms of graphs, there are two intercepts for the graph of the function f(x) = 2x2 + 6x - 5.


Example 9.

4x2 + 4x + 1 = 0

In this example a = 4, b = 4, and c = 1.

There are two things to notice about this example.

  • There is only one solution. In terms of graphs, this means there is only one x-intercept.

  • The solution simplified so that there is no square root involved. This means that the equation could have been solved by factoring. (All quadratic equations can be solved by factoring! What I mean is it could have been solved easily by factoring.)

4x2 + 4x + 1 = 0.

(2x + 1)2 = 0.

x = -1/2.


Example 10.

x2 + x + 1 = 0

a = 1, b = 1, c = 1

Note: There are no real solutions. In terms of graphs, there are no intercepts for the graph of the function f(x) = x2 + x + 1. Thus, the solutions are complex because the graph of y = x2 + x + 1 has no x-intercepts.

 

The expression under the radical in the Quadratic Formula, b2 - 4ac, is called the discriminant of the equation. The last three examples illustrate the three possibilities for quadratic equations.

1. Discriminant > 0. Two real solutions.

2. Discriminant = 0. One real solution.

3. Discriminant < 0. Two complex solutions.

Notes on checking solutions

None of the techniques introduced so far in this section can introduce extraneous solutions. (See example 3 from the Linear Equations and Modeling section.) However, it is still a good idea to check your solutions, because it is very easy to make careless errors while solving equations.

The algebraic method, which consists of substituting the number back into the equation and checking that the resulting statement is true, works well when the solution is "simple", but it is not very practical when the solution involves a radical.

For instance, in our next to last example, 4x2 + 4x + 1 = 0, we found one solution x = -1/2.

The algebraic check looks like

4(-1/2)2 +4(-1/2) + 1 = 0.

4(1/4) - 2 + 1 = 0.

1 - 2 + 1 = 0.

0 = 0. The solution checks.

In the example before that, 2x2 + 6x - 5 = 0, we found two real solutions, x = (-3 ± sqrt(19))/2. It is certainly possible to check this algebraically, but it is not very easy. In this case either a graphical check, or using a calculator for the algebraic check are faster.

First, find decimal approximations for the two proposed solutions.

(-3 + sqrt(19))/2 = 0.679449.

(-3 - sqrt(19))/2 = -3.679449.

Now use a graphing utility to graph y = 2x2 + 6x - 5, and trace the graph to find approximately where the x-intercepts are. If they are close to the values above, then you can be pretty sure you have the correct solutions. You can also insert the approximate solution into the equation to see if both sides of the equation give approximately the same values. However, you still need to be careful in your claim that your solution is correct, since it is not the exact solution.

Note that if you had started with the equation 2x2 + 6x - 5 = 0 and gone directly to the graphing utility to solve it, then you would not get the exact solutions, because they are irrational. However, having found (algebraically) two numbers that you think are solutions, if the graphing utility shows that intercepts are very close to the numbers you found, then you are probably right!

Exercise 1:

Solve the following quadratic equations.

(a) 3x2 -5x - 2 = 0. Answer

(b) (x + 1)2 = 3. Answer

(c) x2 = 3x + 2. Answer

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Equations Involving Radicals

Equations with radicals can often be simplified by raising to the appropriate power, squaring if the radical is a square root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked.

If there is only one radical in the equation, then before raising to a power, you should arrange to have the radical term by itself on one side of the equation.

Example 11.

Now that we have isolated the radical term on the right side, we square both sides and solve the resulting equation for x.

Check:

x = 0

When we substitute x = 0 into the original equation we get the statement 0 = 2, which is not true!

So, x = 0 is not a solution.

x = 3

When we substitute x = 3 into the original equation, we get the statement 3 = 3. This is true, so x = 3 is a solution.

Solution: x = 3.

Note: The solution is the x-coordinate of the intersection point of the graphs of y = x and y = sqrt(x+1)+1.

Look at what would have happened if we had squared both sides of the equation before isolating the radical term.

This is worse than what we started with!

If there is more than one radical term in the equation, then in general, we cannot eliminate all radicals by raising to a power one time. However, we can decrease the number of radical terms by raising to a power.

If the equation involves more than one radical term, then we still want to isolate one radical on one side and raise to a power. Then we repeat that process.

Example 12.

Now square both sides of the equation.

This equation has only one radical term, so we have made progress! Now isolate the radical term and then square both sides again.

Check:

Substituting x = 5/4 into the original equation yields

sqrt(9/4) + sqrt(1/4) = 2.

3/2 + 1/2 = 2.

This statement is true, so x = 5/4 is a solution.

Note on checking solutions:

The algebraic check was easy to do in this case. However, the graphical check has the advantage of showing that there are no solutions that we have not found, at least within the scope of the viewing rectangle. The solution is the x-coordinate of the intersection point of the graphs of y = 2 and y = sqrt(x+1)+sqrt(x-1).

Exercise 2:

Solve the equation sqrt(x+2) + 2 = 2x. Answer

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Polynomial Equations of Higher Degree

We have seen that any degree two polynomial equation (quadratic equation) in one variable can be solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a problem, then either the polynomial is of a special form which allows us to factor it, or we must approximate the solutions with a graphing utility.

Zero Constant

One common special case is where there is no constant term. In this case we may factor out one or more powers of x to begin the problem.

Example 13.

2x3 + 3x2 -5x = 0.

x (2x2 + 3x -5) = 0.

Now we have a product of x and a quadratic polynomial equal to 0, so we have two simpler equations.

x = 0, or 2x2 + 3x -5 = 0.

The first equation is trivial to solve. x = 0 is the only solution. The second equation may be solved by factoring. Note: If we were unable to factor the quadratic in the second equation, then we could have resorted to using the Quadratic Formula. [Verify that you get the same results as below.]

x = 0, or (2x + 5)(x - 1) = 0.

So there are three solutions: x = 0, x = -5/2, x = 1.

Note: The solution is found from the intercepts of the graphs of   f(x) = 2x3 + 3x2 -5x.

Factor by Grouping

Example 14.

x3 -2x2 -9x +18 = 0.

The coefficient of x2 is -2 times that of x3, and the same relationship exists between the coefficients of the third and fourth terms. Group terms one and two, and also terms three and four.

x2 (x - 2) - 9 (x - 2) = 0.

These groups share the common factor (x - 2), so we can factor the left hand side of the equation.

(x - 2)(x2 - 9) = 0.

Whenever we find a product equal to zero, we obtain two simpler equations.

x - 2 = 0, or x2 - 9 = 0.

x = 2, or (x + 3)(x - 3) = 0.

So, there are three solutions, x = 2, x = -3, x = 3.

Note: These solutions are found from the intercepts of the graph of   f(x) = x3 -2x2 -9x +18.

Quadratic in Form

Example 15.

x4 - x2 - 12 = 0.

This polynomial is not quadratic, it has degree four. However, it can be thought of as quadratic in x2.

(x2) 2 -(x2) - 12 = 0.

It might help you to actually substitute z for x2.

z2 - z - 12 = 0 This is a quadratic equation in z.

(z - 4)(z + 3) = 0.

z = 4 or z = -3.

We are not done, because we need to find values of x that make the original equation true. Now replace z by x2 and solve the resulting equations.

x2 = 4.

x = 2, x = -2.

 

x2 = -3.

x = i , or x = -i.

So there are four solutions, two real and two complex.

Note: These solutions are found from the intercepts of the graph of   f(x) = x4 - x2 - 12.

A graph of f(x) = x4 - x2 - 12  and a zoom showing its local extrema.

Exercise 3:

Solve the equation x4 - 5x2 + 4 = 0. Answer

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Equations Involving Fractional Expressions or Absolute Values

Example 16.

The least common denominator is x(x + 2), so we multiply both sides by this product.

This equation is quadratic. The Quadratic Formula yields the solutions

Checking is necessary because we multiplied both sides by a variable expression. Using a graphing utility we see that both of these solutions check. The solution is the x-coordinate of the intersection point of the graphs of y = 1 and y = 2/x-1/(x+2).

Example 17.

5 | x - 1 | = x + 11.

The key to solving an equation with absolute values is to remember that the quantity inside the absolute value bars could be positive or negative. We will have two separate equations representing the different possibilities, and all solutions must be checked.

Case 1. Suppose x - 1 >= 0. Then | x - 1 | = x - 1, so we have the equation

5(x - 1) = x + 11.

5x - 5 = x + 11.

4x = 16.

x = 4, and this solution checks because 5*3 = 4 + 11.

Case 2. Suppose x - 1 < 0. Then x - 1 is negative, so | x - 1 | = -(x - 1). This point often confuses students, because it looks as if we are saying that the absolute value of an expression is negative, but we are not. The expression (x - 1) is already negative, so -(x - 1) is positive.

Now our equation becomes

-5(x - 1) = x + 11.

-5x + 5 = x + 11.

-6x = 6.

x = -1, and this solution checks because 5*2 = -1 + 11.

If you use the Java Grapher to check graphically, note that abs() is absolute value, so you would graph

and look at x-intercepts, or you can find the solution as the x-coordinates of the intersection points of the graphs of y = x+11 and y = 5*abs(x-1).

Exercise 4:

(a) Solve the equation Answer

(b) Solve the equation | x - 2 | = 2 - x/3 Answer

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Sours: http://dl.uncw.edu/digilib/Mathematics/Algebra/mat111hb/IZS/asolve/asolve.html

solve $x^2 = 9 \pmod {100}$

Here is an alternative approach that requires very little theory beyond the basics of integer factorization.

You need to solve the equation $$(x-3)(x+3) \equiv 0 \pmod {100}$$ Interpret this equation this way: $x-3$ and $x+3$ are two numbers that differ by $6$, and whose product is a multiple of $100$. We need to find two such numbers.

Now the two numbers, multiplied together, must have at least two factors of $5$ and two factors of $2$ (because any multiple of $100$ must contain all of those). Because our two unknown numbers differ by exactly $6$, the two factors of $5$ can't be split between the two unknown numbers (if one of them is a multiple of $5$, the other one can't be). So one of the two unknowns must be a multiple of $25$ -- which means it is $25,50$ or $75$. (I'm ignoring the trivial case where one of the numbers is $0$, because that corresponds to the solutions $x = \pm 3$, which you already know about.)

But if one of the numbers is $25$, then the other number is $25 \pm 6$, in which case they are both odd -- and we need to get our two factors of $2$ from somewhere, so that possibility is ruled out. For the same reason, we can rule out the possibility that one of the numbers is $75$.

So we are down to only one possibility: One of the numbers is $50$, and the other number is either $44$ or $56$. Notice that either of these two cases will work: $$ 44 \cdot 50 \equiv 0 \pmod{100}$$ and $$ 50 \cdot 56 \equiv 0 \pmod{100}$$

These two possibilities correspond to the solutions $x =47$ and $x=53$, respectively.

answered Dec 9 '15 at 19:05

mweissmweiss

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$\endgroup$Sours: https://math.stackexchange.com/questions/1567656/solve-x2-9-pmod-100/1567935
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What are the solutions to the quadratic equation x2 9 0?

Solving Quadratics Using Square Roots One way to solve the quadratic equation x2 = 9 is to subtract 9 from both sides to get one side equal to 0: x2 – 9 = 0. The expression on the left can be factored: (x + 3)(x – 3) = 0. Using the zero factor property, you know this means x + 3 = 0 or x – 3 = 0, so x = −3 or 3.

How many solutions does x2 9 have?

The equation has no real solutions. It has 2 imaginary, or complex solutions.

What is the factor of x² 9?

Answer. One way to solve the quadratic equation x2 = 9 is to subtract 9 from both sides to get one side equal to 0: x2 – 9 = 0. The expression on the left can be factored: (x + 3)(x – 3) = 0.

Is X squared +49 prime?

The polynomial x2−49 x 2 – 49 is not prime because the discriminant is a perfect square number.

What are the factors of x squared 49?

Rewrite 49 as 72 . Since both terms are perfect squares, factor using the difference of squares formula, a2−b2=(a+b)(a−b) a 2 – b 2 = ( a + b ) ( a – b ) where a=x and b=7 .

What is the factor of x 2 25?

1 Answer. The answer is (x+5)(x−5) .

What should be multiplied to X 7 to get X² 49?

When you multiply is all the answer is x2-49 because x multiply to x is equal to x2 then – 7 multiply to +7 is equal to – 49 because when you multiply the negative and positive the answer is negative. 2. It is the same in number one you just multiply it. 3.

What is the LCM of 63 and 76?

4788

What is the LCM of 28 and 35?

140

What is the least common multiple of 8 72 216?

  • Step by Step Solution. Least Common Multiple is : 216. Calculate Least Common Multiple for : 8, 72 and 216. Factorize of the above numbers : 8 = 23 72 = 23 • 32 216 = 23 • 33
  • Why learn this.
  • Terms and topics. LCM calculator.
  • Related links. LCM on Purplemath. LCM on Math.com. LCM on The School Run. LCM for Dummies.

2020-10-10Alex SmithMath

Sours: https://rehabilitationrobotics.net/what-are-the-solutions-to-the-quadratic-equation-x2-9-0/
How to Solve Quadratic Equations - Using 3 Different Methods

Square Roots and Completing the Square

 

Learning Objective(s)

·         Solve quadratic equations by using the Square Root Property.

·         Identify and complete perfect square trinomials.

·         Solve quadratic equations by completing the square.

 

Quadratic equations can be solved in many ways. You may already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this topic, you will use square roots to learn another way to solve quadratic equations—and this method will work with all quadratic equations.

 

Solving Quadratics Using Square Roots

One way to solve the quadratic equation x2  = 9 is to subtract 9 from both sides to get one side equal to 0: x2  – 9 = 0. The expression on the left can be factored:

(x + 3)(x – 3) = 0. Using the zero factor property, you know this means x + 3 = 0 or x – 3 = 0, so x = −3 or 3.

 

Another property would let you solve that equation more easily.

 

TheSquare Root Property

 

If x2 = a, then x =  or .

 

The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of a and the negative square root of a.

 

 

A shortcut way to write “” or “” is . The symbol ± is often read “positive or negative.” If it is used as an operation (addition or subtraction), it is read “plus or minus.”

 

 

Example

Problem

Solve using the Square Root Property. x2 = 9

 

x2 = 9

x =

x = ±3

Since one side is simply x2, you can take the square root of both sides to get x on one side. Don’t forget to use both positive and negative square roots!

Answer

x = ±3 (that is, x = 3 or −3)

 

 

Notice that there is a difference here in solving x2 = 9 and finding . For x2 = 9, you are looking for all numbers whose square is 9. For , you only want the principal (nonnegative) square root. The negative of the principal square root is ; both would be . Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted!

 

In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating x2.

 

 

Example

Problem

Solve. 10x2 + 5 = 85

 

 

10x2 + 5 = 85

 

 

 

 

 

 

If you try taking the square root of both sides of the original equation, you will have  on the left, and you can’t simplify that. Subtract 5 from both sides to get the x2 term by itself.

 

10x2 = 80

 

You could now take the square root of both sides, but you would have

as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.

 

Now you have only x2 on the left, so you can use the Square Root Property easily.

 

Be sure to simplify the radical if possible.

 

 

Answer

 

 

 

Sometimes more than just the x is being squared:

 

 

Example

Problem

Solve. (x – 2)2 – 50 = 0

 

 

(x – 2)2 – 50 = 0

Again, taking the square root of both sides at this stage will leave something you can’t work with on the left. Start by adding 50 to both sides.

 

(x – 2)2 = 50

x – 2 =

Because (x – 2)2 is a squared quantity, you can take the square root of both sides.

 

To isolate x on the left, you need to add 2 to both sides.

 

Be sure to simplify the radical if possible.

 

 

Answer

 

 

 

This method can be helpful when solving real-world problems.

 

 

Example

Problem

The formula for compounding interest annually is

A = P(1 + r)t, where A is the balance after t years, when P is the principal (initial amount invested) and r is the interest rate.

 

Find the interest rate r if $3,000 is invested and grows to $3,307.50 after 2 years.

 

A= P(1 + r)t

A = 3,307.50

t = 2

P = 3,000

First identify what you know. The amount after 2 years is 3,307.50, so

A = 3,307.50. This also means t = 2. The principal P is the original amount invested, so that is 3,000.

 

3,307.50 = 3,000(1 + r)2

Substitute the values for the variables you know. Only r is left, so try to isolate r.

 

Dividing both sides by 3000 leaves only (1 + r)2 on the right. Because (1 + r)2 is a squared quantity you can use the Square Root Property.

 

Don’t forget the ±!

 

±1.05 = 1 + r

Using a calculator, you can find that  is 1.05.

 

±1.05 – 1 = r

Subtract 1 from both sides to isolate r on the right.

 

1.05 – 1 = r, or

−1.05 – 1 = r

You now have two equations, one using 1.05 and one using −1.05.

 

r = 0.05 or −2.05

Simplifying the two equations gives two solutions to the equation.

Answer

The interest rate is 0.05, or 5%.

Notice that a negative interest rate doesn’t make sense for this context, so only the positive value could be the interest rate. The -2.05 is an extraneous solution and must be discarded.

 

 

Solve. (x – 3)2 – 2 = 16

 

A) x = 3 ±

 

B) x = 3 +

 

C) x = 7

 

D) x = 1 or 9

 

Show/Hide Answer

A) x = 3 ±

Correct. Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives x – 3 = , so x = 3 ± . Simplifying the radical gives .

 

B) x = 3 +

Incorrect. You forgot the negative square root when you took the square root of both sides. Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives x – 3 = , so x = 3 ± . Simplifying the radical gives 3 ± .

 

C) x = 7

Incorrect. There are two mistakes here: Knowing the square root of 16 may have made you forget that to solve this equation, the squared quantity needs to be isolated Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives

x – 3 = . (Note that both the positive and negative square roots are included; this is the other probable mistake.) So, x = 3 ± . Simplifying the radical gives 3 ± .

 

D) x = 1 or 9

Incorrect. Knowing the square root of 16 may have made you forget that to solve this equation, the squared quantity needs to be isolated Before taking the square root, add 2 to both sides: (x – 3)2 = 18. Applying the Square Root Property gives x – 3 = . So, x = 3 ± . Simplifying the radical gives 3 ± .

 

 

 

Perfect Square Trinomials

Of course, quadratic equations often will not come in the format of the examples above. Most of them will have x terms. However, you may be able to factor the expression into a squared binomial—and if not, you can still use squared binomials to help you.

 

First, let’s look at squared binomials. Some of the above examples have squared binomials: (1 + r)2 and (x – 2)2 are squared binomials. (They are binomials, two terms, that are squared.) If you expand these, you get a perfect square trinomial. For example, (1 + r)2 = (1 + r)(1 + r) = 1 + 2r + r2, or r2 + 2r + 1. The trinomial r2 + 2r + 1 is a perfect square trinomial. Notice that the first and last terms are squares (r2 and 1). The middle term is twice the product of the square roots of the first and last terms, the square roots are r and 1, and the middle term is 2(r)(1).

 

Perfect square trinomials have the form r2 + 2rs + s2 and can be factored as (r + s)2, or they have the form r2 – 2rs + s2 and can be factored as (rs)2. Let’s factor a perfect square trinomial into a squared binomial.

 

 

Example

Problem

Factor 9x2 – 24x + 16.

 

 

9x2 = (3x)2

16 = 42

First notice that the x2 term and the constant term are both perfect squares.

 

24x = 2(3x)(4)

 

Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms.

 

r = 3x

s = 4

 

9x2 – 24x + 16 = (3x – 4)2

A trinomial in the form r2 – 2rs + s2 can be factored as (rs)2.

 

In this case, the middle term is subtracted, so subtract r and s and square it to get (rs)2.

Answer

(3x – 4)2

 

 

 

You can use the procedure in this next example to help you solve equations where you identify perfect square trinomials, even if the equation is not set equal to 0.

 

 

Example

Problem

Solve. 4x2 + 20x + 25 = 8

 

 

4x2 + 20x + 25 = 8

Since there’s an x term, you can’t use the Square Root Property immediately (or even after adding or dividing by a constant).

 

Notice, however, that the x2 and constant terms on the left are both perfect squares: (2x)2 and 52. Check the middle term: is it 2(2x)(5)? Yes!

 

(2x + 5)2 = 8

 

A trinomial in the form r2 + 2rs + s2can be factored as (r + s)2, so rewrite the left side as a squared binomial.

 

Now you can use the Square Root Property. Some additional steps are needed to isolate x.

 

Simplify the radical when possible.

 

Answer

 

 

 

 

You may have looked at the problem above and thought “why not subtract 8 from both sides first, making the equation 4x2 + 20x + 17 = 0?” This is one technique for solving for x—the problem is that even if you did this, the equation 4x2 + 20x + 17 = 0 cannot be factored with real numbers. (Try it—can you think of two numbers whose product is 68 and whose sum is 20?)

 

Since the expression 4x2 + 20x + 25 can be identified as a perfect square trinomial, it is best to factor it as (2x + 5)2 and then use the Square Root Property.

 

One way to solve quadratic equations is by completing the square. When you don’t have a perfect square trinomial, you can create one byadding a constant term that is a perfect square to both sides of the equation. Let’s see how to find that constant term.

 

“Completing the square” does exactly what it says—it takes something that is not a square and makes it one. This idea can be illustrated using an area model of the binomial x2 + bx.

 

 

In this example, the area of the overall rectangle is given by x(x + b).

 

Now let's make this rectangle into a square. First, divide the red rectangle with area bx into two equal rectangles each with area . Then rotate and reposition one of them. You haven't changed the size of the red area—it still adds up to bx.

 

 

 

The red rectangles now make up two sides of a square, shown in white. The area of that square is the length of the red rectangles squared, or .

 

Here comes the cool part—do you see that when the white square is added to the blue and red regions, the whole shape is also now a square? In other words, you've "completed the square!" By adding the quantity  to the original binomial, you've made a square, a square with sides of .

 

Notice that the area of this square can be written as a squared binomial: .

 

 

Finding a Value that will Complete the Square in an Expression

 

To complete the square for an expression of the form x2 + bx:

·         Identify the value of b;

·         Calculate and add .

The expression becomes .

 

 

 

Example

Problem

Find the number to add to x2 + 8x to make it a perfect square trinomial.

 

x2+ 8x

b = 8

 

First identify b if this has the form x2+ bx.

 

 

 

 

To complete the square, add .

 

b = 8, so

 

x2 + 8x + (4)2

x2 + 8x + 16

 

x2 + 8x + 16 = (x + 4)2

 

Simplify.

 

Check that the result is a perfect square trinomial. (x + 4)2 =

x2 + 4x + 4x + 16 =

x2 + 8x + 16, so it is.

Answer

Adding +16 will make x2 + 8x a perfect square trinomial.

 

 

Notice that  is always positive, since it is the square of a number. When you complete the square, you are always adding a positive value.

 

 

Use completing the square to find the value to add that makes x2 – 12x a perfect square trinomial. Then write the expression as the square of a binomial.

 

A) add 12; (x – 6)2

B) add 36; (x + 6)2

C) add −12; (x – 12)2

D) add 36; (x – 6)2

 

Show/Hide Answer

A) add 12; (x – 6)2

Incorrect. The value of b is −12, sothe value to add is , not 12. Add 36 to get x2 – 12x + 36. The correct answer is (x – 6)2.

 

B) add 36; (x + 6)2

Incorrect. The value to add has been calculated correctly: . However, the resulting trinomial is x2 – 12x + 36, which factors to (x – 6)2, not (x + 6)2. The correct answer is (x – 6)2.

 

C) add −12; (x – 12)2

Incorrect. The value of b is −12, so the value to add is , not −12. Note also, that the number you add will always be positive because it is the square of a number. The correct answer is (x – 6)2.

 

D) add 36; (x – 6)2

Correct. The value to add is  and the resulting trinomial x2 – 12x + 36 factors to (x – 6)2.

 

 

 

Solving a Quadratic Equation using Completing the Square

 

You can use completing the square to help you solve a quadratic equation that cannot be solved by factoring.

 

Let’s start by seeing what happens when you complete the square in an equation. In the example below, notice that completing the square will result in adding a number to both sides of the equation—you have to do this in order to keep both sides equal!

 

 

Example

Problem

Rewrite x2 + 6x = 8 so that the left side is a perfect square trinomial.

 

x2 + 6x = 8

b = 6

 

This equation has a constant of 8. Ignore it for now and focus on the x2 and x terms on the left side of the equation. The left side has the form x2+ bx, so you can identify b.

 

 

x2 + 6x+ 9 = 8 + 9

 

To complete the square, add  to the left side.

 

b = 6, so

 

This is an equation, though, so you must add the same number to the right side as well.

 

 

x2 + 6x + 9 = 17

 

x2 + 6x + 9 = 17

(x + 3)2 = 17

 

Simplify.

 

Check that the left side is a perfect square trinomial.

(x + 3)2 =

x2 + 3x + 3x + 9 =

x2 + 6x + 9, so it is.

Answer

x2 + 6x + 9 = 17

 

 

 

Can you see that completing the square in an equation is very similar to completing the square in an expression? The main difference is that you have to add the new number (+9 in this case) to both sides of the equation to maintain equality.

 

Now let’s look at an example where you are using completing the square to actually solve an equation, finding a value for the variable.

 

 

Example

Problem

Solve. x2 – 12x – 4 = 0

 

 

x2 – 12x = 4

b = −12

 

 

 

Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.

 

Rewrite the equation with the left side in the form
x2 + bx, to prepare to complete the square. Identify b.

 

 

 

x2 – 12x+ 36 = 4 + 36

 

x2 – 12x + 36 = 40

 

Figure out what value to add to complete the square. Add  to complete the square, so = .

 

Add the value to both sides of the equation and simplify.

 

(x – 6)2 =  40

 

 

Rewrite the left side as a squared binomial.

 

 

 

 

Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.

 

 

Solve for x by adding 6 to both sides. Simplify as needed.

 

 

Answer

 

 

 

You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that’s slightly different.

 

 

Example

Problem

Solve. x2 + 16x + 17 = −47

 

x2 + 16x = −64

b = 16

 

Rewrite the equation so the left side has the form x2 + bx. Identify b.

 

 

x2 + 16x + 64 = −64 + 64

x2 + 16x + 64 = 0

 

Add , which is , to both sides.

 

 

(x + 8)2 = 0

 

Write the left side as a squared binomial.

 

x + 8 = 0

 

 

 

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.

 

x = −8

 

Solve for x.

Answer

x = −8

 

 

 

Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 47 to both sides in the equation. The equation x2 + 16x + 17 = −47 becomes x2 + 16x + 64 = 0. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 16).

 

It can be factored as (x + 8)(x + 8) = 0, of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.

 

 

Solve. x2 – 16x = −1

 

A)

B)

C)

D)

 

Show/Hide Answer

A)

Incorrect. You appear to have added 16 and then factored the trinomial incorrectly. However, the value of b is −16, so to complete the square, add  (not 16). Add 64 to both sides to get x2 – 16x + 64 = 63. This is (x – 8)2 = 63, so  That means  The correct answer is .

 

B)

Incorrect. The value to add has been calculated correctly:  However, the resulting trinomial is x2 – 16x + 64, which factors to (x – 8)2, not (x + 8)2. The equation becomes (x – 8)2 = 63, so  That means  The correct answer is .

 

C)

Incorrect. The value to add has been calculated correctly:  However, you appear to have added 64 to the left side of the equation only, and not to the right side. The equation becomes (x – 8)2 = 63, so  That means  The correct answer is .

 

D)

Correct. The value of b is −16, so to complete the square, add . Add 64 to both sides to get x2 – 16x + 64 = 63. This is (x – 8)2 = 63, so  That means

 

 

 

Completing the square is used to change a binomial of the form x2 + bx into a perfect square trinomial , which can be factored to . When solving quadratic equations by completing the square, be careful to add  to both sides of the equation to maintain equality. The Square Root Property can then be used to solve for x. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.

 

Sours: http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U16_L5_T1_text_final.html

X2 9 solve

Simplifying radicals

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x2"   was replaced by   "x^2". 

Step by step solution :

Step  1  :

Polynomial Roots Calculator :

 1.1    Find roots (zeroes) of :       F(x) = x2+9
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  9.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,3 ,9

 Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor
     -1     1      -1.00      10.00   
     -3     1      -3.00      18.00   
     -9     1      -9.00      90.00   
     1     1      1.00      10.00   
     3     1      3.00      18.00   
     9     1      9.00      90.00   


Polynomial Roots Calculator found no rational roots

Equation at the end of step  1  :

x2 + 9 = 0

Step  2  :

Solving a Single Variable Equation :

 2.1      Solve  :    x2+9 = 0 

 Subtract  9  from both sides of the equation : 
                      x2 = -9
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ -9  

 In Math,  i  is called the imaginary unit. It satisfies   i2  =-1. Both   i   and   -i   are the square roots of   -1 

Accordingly,  √ -9  =
                    √ -1• 9   =
                    √ -1 •√ 9   =
                    i •  √ 9

Can  √ 9 be simplified ?

Yes!   The prime factorization of  9   is
   3•3 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 9   =  √ 3•3  =
                ±  3 • √ 1   =
                ±  3

The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      x=  0.0000 + 3.0000 i 
                      x=  0.0000 - 3.0000 i 

Two solutions were found :

  1.   x=  0.0000 - 3.0000 i 
  2.   x=  0.0000 + 3.0000 i 
Sours: https://www.tiger-algebra.com/drill/x2_9=0/
Does x2 + y2 = 9 Represent y as a Function of x?

Extracting Square Roots

Recall that a quadratic equation is in standard formAny quadratic equation in the form , where a, b, and c are real numbers and . if it is equal to 0:

where a, b, and c are real numbers and . A solution to such an equation is called a rootA solution to a quadratic equation in standard form.. Quadratic equations can have two real solutions, one real solution, or no real solution. If the quadratic expression on the left factors, then we can solve it by factoring. A review of the steps used to solve by factoring follow:

Step 1: Express the quadratic equation in standard form.

Step 2: Factor the quadratic expression.

Step 3: Apply the zero-product property and set each variable factor equal to 0.

Step 4: Solve the resulting linear equations.

For example, we can solve by factoring as follows:

The two solutions are −2 and 2. The goal in this section is to develop an alternative method that can be used to easily solve equations where b = 0, giving the form

The equation is in this form and can be solved by first isolating .

If we take the square root of both sides of this equation, we obtain the following:

Here we see that and are solutions to the resulting equation. In general, this describes the square root propertyFor any real number k, if , then .; for any real number k,

The notation “±” is read “plus or minus” and is used as compact notation that indicates two solutions. Hence the statement indicates that or . Applying the square root property as a means of solving a quadratic equation is called extracting the rootsApplying the square root property as a means of solving a quadratic equation..

Example 1: Solve: .

Solution: Begin by isolating the square.

Next, apply the square root property.

Answer: The solutions are −5 and 5. The check is left to the reader.

Certainly, the previous example could have been solved just as easily by factoring. However, it demonstrates a technique that can be used to solve equations in this form that do not factor.

Example 2: Solve: .

Solution: Notice that the quadratic expression on the left does not factor. We can extract the roots if we first isolate the leading term, .

Apply the square root property.

For completeness, check that these two real solutions solve the original quadratic equation. Generally, the check is optional.

Answer: The solutions are and .

Example 3: Solve: .

Solution: Begin by isolating .

Apply the square root property and then simplify.

Answer: The solutions are and .

Sometimes quadratic equations have no real solution.

Example 4: Solve: .

Solution: Begin by isolating .

After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation.

Answer: No real solution

Reverse this process to find equations with given solutions of the form ±k.

Example 5: Find an equation with solutions and .

Solution: Begin by squaring both sides of the following equation:

Lastly, subtract 12 from both sides and present the equation in standard form.

Answer:

Try this! Solve: .

Answer: or

Video Solution

(click to see video)

Consider solving the following equation:

To solve this equation by factoring, first square and then put it in standard form, equal to zero, by subtracting 25 from both sides.

Factor and then apply the zero-product property.

The two solutions are −7 and 3.

When an equation is in this form, we can obtain the solutions in fewer steps by extracting the roots.

Example 6: Solve: .

Solution: Solve by extracting the roots.

At this point, separate the “plus or minus” into two equations and simplify each individually.

Answer: The solutions are −7 and 3.

In addition to fewer steps, this method allows us to solve equations that do not factor.

Example 7: Solve: .

Solution: Begin by isolating the square.

Next, extract the roots and simplify.

Solve for x.

Answer: The solutions are and .

Example 8: Solve: .

Solution: Begin by isolating the square factor.

Apply the square root property and solve.

Answer: The solutions are and .

Try this! Solve: .

Answer:

Video Solution

(click to see video)

Example 9: The length of a rectangle is twice its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle.

Solution:

The diagonal of any rectangle forms two right triangles. Thus the Pythagorean theorem applies. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse:

Solve.

Here we obtain two solutions, and . Since the problem asked for a length of a rectangle, we disregard the negative answer. Furthermore, we will rationalize the denominator and present our solutions without any radicals in the denominator.

Back substitute to find the length.

Answer: The length of the rectangle is feet and the width is feet.

Key Takeaways

  • Solve equations of the form by extracting the roots.
  • Extracting roots involves isolating the square and then applying the square root property. After applying the square root property, you have two linear equations that each can be solved. Be sure to simplify all radical expressions and rationalize the denominator if necessary.

Topic Exercises

Part A: Extracting Square Roots

Solve by factoring and then solve by extracting roots. Check answers.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

Solve by extracting the roots.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

Find a quadratic equation in standard form with the following solutions.

63. ±7

64. ±13

65.

66.

67.

68.

69.

70.

Solve and round off the solutions to the nearest hundredth.

71.

72.

73.

74.

75.

76.

77.

78.

Set up an algebraic equation and use it to solve the following.

79. If 9 is subtracted from 4 times the square of a number, then the result is 3. Find the number.

80. If 20 is subtracted from the square of a number, then the result is 4. Find the number.

81. If 1 is added to 3 times the square of a number, then the result is 2. Find the number.

82. If 3 is added to 2 times the square of a number, then the result is 12. Find the number.

83. If a square has an area of 8 square centimeters, then find the length of each side.

84. If a circle has an area of square centimeters, then find the length of the radius.

85. The volume of a right circular cone is cubic centimeters when the height is 6 centimeters. Find the radius of the cone. (The volume of a right circular cone is given by .)

86. The surface area of a sphere is square centimeters. Find the radius of the sphere. (The surface area of a sphere is given by .)

87. The length of a rectangle is 6 times its width. If the area is 96 square inches, then find the dimensions of the rectangle.

88. The base of a triangle is twice its height. If the area is 16 square centimeters, then find the length of its base.

89. A square has an area of 36 square units. By what equal amount will the sides have to be increased to create a square with double the given area?

90. A circle has an area of square units. By what amount will the radius have to be increased to create a circle with double the given area?

91. If the sides of a square measure 1 unit, then find the length of the diagonal.

92. If the sides of a square measure 2 units, then find the length of the diagonal.

93. The diagonal of a square measures 5 inches. Find the length of each side.

94. The diagonal of a square measures 3 inches. Find the length of each side.

95. The length of a rectangle is twice its width. If the diagonal measures 10 feet, then find the dimensions of the rectangle.

96. The length of a rectangle is twice its width. If the diagonal measures 8 feet, then find the dimensions of the rectangle.

97. The length of a rectangle is 3 times its width. If the diagonal measures 5 meters, then find the dimensions of the rectangle.

98. The length of a rectangle is 3 times its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle.

99. The height in feet of an object dropped from a 9‑foot ladder is given by , where t represents the time in seconds after the object has been dropped. How long does it take the object to hit the ground? (Hint: The height is 0 when the object hits the ground.)

100. The height in feet of an object dropped from a 20‑foot platform is given by , where t represents the time in seconds after the object has been dropped. How long does it take the object to hit the ground?

101. The height in feet of an object dropped from the top of a 144-foot building is given by , where t is measured in seconds.

a. How long will it take to reach half of the distance to the ground, 72 feet?

b. How long will it take to travel the rest of the distance to the ground?

Round off to the nearest hundredth of a second.

102. The height in feet of an object dropped from an airplane at 1,600 feet is given by , where t is in seconds.

a. How long will it take to reach half of the distance to the ground?

b. How long will it take to travel the rest of the distance to the ground?

Round off to the nearest hundredth of a second.

Part B: Discussion Board

103. Create an equation of your own that can be solved by extracting the root. Share it, along with the solution, on the discussion board.

104. Explain why the technique of extracting roots greatly expands our ability to solve quadratic equations.

105. Explain in your own words how to solve by extracting the roots.

106. Derive a formula for the diagonal of a square in terms of its sides.

Answers

1: −6, 6

3: −3/2, 3/2

5: 1, 3

7: 1/2, 7/2

9: −1, 3

11: 0, 10

13: ±4

15: ±3

17: ±1/2

19: ±0.5

21:

23: ±3/4

25:

27: ±10

29: No real solution

31: ±2/3

33: ±0.3

35:

37:

39: No real solution

41:

43:

45:

47: −9, −5

49: −3, 6

51:

53:

55:

57:

59: No real solution

61:

63:

65:

67:

69:

71: ±0.33

73: ±5.66

75: ±7.94

77: ±3.61

79: or

81: or

83: centimeters

85: centimeters

87: Length: 24 inches; width: 4 inches

89: units

91: units

93: inches

95: Length: feet; width: feet

97: Length: meters; width: meters

99: 3/4 second

101: a. 2.12 seconds; b. 0.88 second

Sours: https://saylordotorg.github.io/text_elementary-algebra/s12-01-extracting-square-roots.html

You will also be interested:

What are the roots of the quadratic equation X2 9?

Solving Quadratics Using Square Roots One way to solve the quadratic equation x2 = 9 is to subtract 9 from both sides to get one side equal to 0: x2 – 9 = 0. The expression on the left can be factored: (x + 3)(x – 3) = 0. Using the zero factor property, you know this means x + 3 = 0 or x – 3 = 0, so x = −3 or 3.

What is the discriminant of X² 6x 9?

0

Which is a quadratic equation?

A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.

What do you call the expression b2 4ac?

The expression b2 – 4ac is called the discriminant. All quadratic equations have two roots/solutions. These roots are either REAL, EQUAL or COMPLEX.

How important is the expression b2-4ac?

what do you think is the importance of the expressions b2-4ac in determining the nature of the roots of quadratic equation? it is very important so we can identify its discriminant or nature of roots whether it is real solution or equal, not equal, rational, irrational.

What is the value of the expression b2-4ac?

The value of the expression b2-4ac is called the discriminant of the quadratic equation ax2+bx+c=0. This value can be used to describe the nature of the roots of. a quadratic equation. It can be zero, positive and perfect square, positive but not.

How many solutions if the discriminant is less than 0?

It tells you the number of solutions to a quadratic equation. If the discriminant is greater than zero, there are two solutions. If the discriminant is less than zero, there are no solutions and if the discriminant is equal to zero, there is one solution.

Under what condition will ax2 5x 7 0 be a quadratic equation?

Explanation: Based on the quadratic formula x=−b±√b2−4ac2a and the form ax2+bx+c=0, we see that a=1, b=5 and c=7. With i=√−1, x=−5±√3i2. Thus, the roots of the equation are x=−5+√3i2 and x=−5−√3i2.

What is the nature of the roots of 3×2 5x 2 0?

If D is equal to 0, then we get two roots which are equal and same. If D is less than 0, then we get roots which are imaginary or unreal. Since D is greater than 0 in this case, we get two real and distinct roots. Hence solved !!

Sours: https://answerstoall.com/object/what-are-the-roots-of-the-quadratic-equation-x2-9/


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