solving quadratic equations completing the square (continued) assignment quizlet

Solving Quadratics by Completing the Square

How to solve quadratic equations using the completing the square method.

If you are already familiar with the steps involved in completing the square , you may skip the introductory discussion and review the seven (7) worked examples right away.

The key step in this method is to find the constant “[latex]k[/latex]” that will allow us to express the given trinomial as the square of a binomial.

For instance,

The value of “[latex]k[/latex]” is determined by squaring one-half of the coefficient of [latex]x[/latex]. In this case, the coefficient of the linear term [latex]x[/latex] is [latex] – \,6[/latex].

Therefore, half of [latex] – \,6[/latex] is [latex] – \,3[/latex], and its square [latex]{\left( { – \,3} \right)^2} = 9[/latex]. The value of [latex]k[/latex] must be [latex]9[/latex]!

If I substitute [latex]k[/latex] by [latex]9[/latex], the trinomial is factored into two equal binomials. This is great because I can now rewrite it in a more compact form, that is, the square of a single binomial.

The following are the general steps involved in solving quadratic equations using completing the square method.

Key Steps in Solving Quadratic Equation by Completing the Square

1) Keep all the [latex]x[/latex]-terms (both the squared and linear) on the left side, while moving the constant to the right side.

In symbol, rewrite the general form [latex]a{x^2} + bx + c[/latex] as:

[latex]a{x^2} + bx = – \,c[/latex]

2) Now, identify what type of problem you have by looking at the coefficient of the leading term, [latex]a[/latex].

• “Easy Type” when [latex]a = 1[/latex]

• “Difficult Type” when [latex]a \ne 1[/latex]

3) If you have the “Easy type”, proceed immediately to Step 4. If you have the “Difficult Type”, you must divide the entire equation first by the value of [latex]a[/latex] before moving to step 4.

4) Take the coefficient of [latex]x[/latex]-term, divide it by [latex]2[/latex] then square its result. Add this value to both sides of the equation.

5) Express the left side as a square of binomial.

6) Get the square root of both sides of the equation. Don’t forget to attach the [latex] \pm [/latex] symbol on the right side!

7) Finish it off by solving the linear equation(s) that arise from it.

Examples of How to Solve Quadratic Equations by Completing the Square

Example 1 : Solve the quadratic equation below by completing the square method.

This is an “Easy Type” since [latex]a = 1[/latex]. I will keep the “[latex]x[/latex]-terms” (both the squared and linear terms) on the left side but move the constant to the right side.

I can do that by adding [latex]15[/latex] on both sides of the equation.

Now, take the coefficient of the linear term (which is the [latex]x[/latex]-term with power [latex]1[/latex]) and perform TWO operations on it:

• Divide by [latex]2[/latex], followed by
• Squaring (raising to the 2nd power)

The output here, which is [latex]+1[/latex], will be added to both sides of the quadratic equation.

This step forces the left side to generate a perfect-square-trinomial which can be expressed as a square of a binomial . Great!

At this point, it is very easy to solve for [latex]x[/latex]. To get rid of the exponent [latex]2[/latex] in the binomial, I will apply square root operation on both sides of the equation.

Next, solve the pair of linear equations that arise as a result of squaring both sides.

Break [latex]x = \pm \,4 + 1[/latex] into two cases, then solve.

That is it! Our answers are [latex]{x_1} = 5[/latex] and [latex]{x_2} = – \,3[/latex].

Make it a habit to check your solved values of [latex]x[/latex] back into the original equation to verify if indeed they are “true” answers. I will leave it to you as an exercise.

Example 2 : Solve the quadratic equation below by completing the square method.

Obviously, I can’t proceed with the steps required in completing the square. I must isolate the [latex]x[/latex]-terms to the left, and the constant to the right.

Do that by subtracting both sides by [latex]1[/latex].

This time I am ready to perform the completing the square steps to solve this quadratic equation. Start by taking the coefficient of the linear [latex]x[/latex]-term then divide it by [latex]2[/latex] followed by squaring it. This is the MOST important step of this whole process.

Whatever number that comes out will be added to both sides of the equation. The left side becomes a perfect square trinomial which can be rewritten as the square of binomial.

Eliminate the power [latex]2[/latex] of the binomial by taking the square root of both sides. I hope that you’ll be able to follow the rest of the solution.

I got the following as the answers which are [latex]{x_1} = 7[/latex] and [latex]{x_2} = 3[/latex]. Go ahead and check the solutions yourself as an exercise.

Example 3 : Solve the quadratic equation below by completing the square method.

The answers are [latex]{x_1} = 2[/latex] and [latex]{x_2} = – 10[/latex]

Example 4 : Solve the quadratic equation below by completing the square method.

The first thing to do is to move the constant to the right side by subtracting each side by [latex]8[/latex].

This is actually the “Difficult Type” since [latex]a \ne 1[/latex]. Thus, I need to make the coefficient of the squared [latex]x[/latex]-term equal to [latex]1[/latex]. This can be done by dividing through the entire equation by [latex]a[/latex] which equals [latex]8[/latex]!

By dividing through by [latex]8[/latex], I have converted this problem into the “easy” case because the coefficient of the squared [latex]x[/latex]-term becomes [latex]+1[/latex]. Finish this off by doing the same process as seen in examples 1 and 2. The only difference is that I will deal with fractions.

Consider the coefficient of the linear [latex]x[/latex]-term, divide by [latex]2[/latex] and square it.

Take the output of the step above, and add to both sides of the quadratic equation. Then proceed with the rest of the steps to complete the square.

The answers should be [latex]{x_1} = 2[/latex] and [latex]{x_2} = {1 \over 2}[/latex].

Example 5  ( Practice Problem ): Solve the quadratic equation below by completing the square method.

Try solving this problem on your own first. Then click below to view the solution.

NOTE: The solution to this problem may look messy but as long as you apply the correct procedures for completing the square, you will soon realize that the answers to this problem come out nicely.

Hint : The solution set includes a rational number and a negative integer.

Example 6 : Solve the quadratic equation below by completing the square method.

I will move the constant to the right side while keeping all [latex]x[/latex]-terms on the left. Then I must divide the entire equation by [latex] – \,3[/latex] since [latex]a \ne 1[/latex].

• Subtract both sides by [latex]42[/latex]

• Divide entire equation by [latex] – \,3[/latex]

Now, I will take the coefficient of the linear term, divide it by [latex]2[/latex] and square it.

Add this output [latex]4[/latex] to both sides of the equation. This makes the left side a perfect-square-trinomial which can be rewritten as the square of a binomial.

That was easy, right? Again, the more you see how these problems are being solved correctly, the better you become!

Example 7  ( Practice Problem ): Solve the quadratic equation below using the completing the square method.

You might also like these tutorials:

• Solving Quadratic Equations by Square Root Method
• Solving Quadratic Equations by Factoring Method
• Solving Quadratic Equations by the Quadratic Formula

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• High School Math Solutions – Quadratic Equations Calculator, Part 2 Solving quadratics by factorizing (link to previous post) usually works just fine. But what if the quadratic equation...

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9.3 Solve Quadratic Equations Using the Quadratic Formula

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic equations using the Quadratic Formula
• Use the discriminant to predict the number and type of solutions of a quadratic equation
• Identify the most appropriate method to use to solve a quadratic equation

Be Prepared 9.7

Before you get started, take this readiness quiz.

Evaluate b 2 − 4 a b b 2 − 4 a b when a = 3 a = 3 and b = −2 . b = −2 . If you missed this problem, review Example 1.21 .

Be Prepared 9.8

Simplify: 108 . 108 . If you missed this problem, review Example 8.13 .

Be Prepared 9.9

Simplify: 50 . 50 . If you missed this problem, review Example 8.76 .

Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for x.

We start with the standard form of a quadratic equation and solve it for x by completing the square.

• Quadratic Formula

The solutions to a quadratic equation of the form ax 2 + bx + c = 0, where a ≠ 0 a ≠ 0 are given by the formula:

To use the Quadratic Formula , we substitute the values of a , b , and c from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.

Notice the formula is an equation. Make sure you use both sides of the equation.

Example 9.21

How to solve a quadratic equation using the quadratic formula.

Solve by using the Quadratic Formula: 2 x 2 + 9 x − 5 = 0 . 2 x 2 + 9 x − 5 = 0 .

Try It 9.41

Solve by using the Quadratic Formula: 3 y 2 − 5 y + 2 = 0 3 y 2 − 5 y + 2 = 0 .

Try It 9.42

Solve by using the Quadratic Formula: 4 z 2 + 2 z − 6 = 0 4 z 2 + 2 z − 6 = 0 .

Solve a quadratic equation using the quadratic formula.

• Step 1. Write the quadratic equation in standard form, ax 2 + bx + c = 0. Identify the values of a , b , and c .
• Step 2. Write the Quadratic Formula. Then substitute in the values of a , b , and c .
• Step 3. Simplify.
• Step 4. Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “ x =”.

Example 9.22

Solve by using the Quadratic Formula: x 2 − 6 x = −5 . x 2 − 6 x = −5 .

Try It 9.43

Solve by using the Quadratic Formula: a 2 − 2 a = 15 a 2 − 2 a = 15 .

Try It 9.44

Solve by using the Quadratic Formula: b 2 + 24 = −10 b b 2 + 24 = −10 b .

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula . If we get a radical as a solution, the final answer must have the radical in its simplified form.

Example 9.23

Solve by using the Quadratic Formula: 2 x 2 + 10 x + 11 = 0 . 2 x 2 + 10 x + 11 = 0 .

Try It 9.45

Solve by using the Quadratic Formula: 3 m 2 + 12 m + 7 = 0 3 m 2 + 12 m + 7 = 0 .

Try It 9.46

Solve by using the Quadratic Formula: 5 n 2 + 4 n − 4 = 0 5 n 2 + 4 n − 4 = 0 .

When we substitute a , b , and c into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

Example 9.24

Solve by using the Quadratic Formula: 3 p 2 + 2 p + 9 = 0 . 3 p 2 + 2 p + 9 = 0 .

Try It 9.47

Solve by using the Quadratic Formula: 4 a 2 − 2 a + 8 = 0 4 a 2 − 2 a + 8 = 0 .

Try It 9.48

Solve by using the Quadratic Formula: 5 b 2 + 2 b + 4 = 0 5 b 2 + 2 b + 4 = 0 .

Remember, to use the Quadratic Formula, the equation must be written in standard form, ax 2 + bx + c = 0. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example 9.25

Solve by using the Quadratic Formula: x ( x + 6 ) + 4 = 0 . x ( x + 6 ) + 4 = 0 .

Our first step is to get the equation in standard form.

Try It 9.49

Solve by using the Quadratic Formula: x ( x + 2 ) − 5 = 0 . x ( x + 2 ) − 5 = 0 .

Try It 9.50

Solve by using the Quadratic Formula: 3 y ( y − 2 ) − 3 = 0 . 3 y ( y − 2 ) − 3 = 0 .

When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.

Example 9.26

Solve by using the Quadratic Formula: 1 2 u 2 + 2 3 u = 1 3 . 1 2 u 2 + 2 3 u = 1 3 .

Our first step is to clear the fractions.

Try It 9.51

Solve by using the Quadratic Formula: 1 4 c 2 − 1 3 c = 1 12 1 4 c 2 − 1 3 c = 1 12 .

Try It 9.52

Solve by using the Quadratic Formula: 1 9 d 2 − 1 2 d = − 1 3 1 9 d 2 − 1 2 d = − 1 3 .

Think about the equation ( x − 3) 2 = 0. We know from the Zero Product Property that this equation has only one solution, x = 3.

We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to 0 gives just one solution. Notice that once the radicand is simplified it becomes 0 , which leads to only one solution.

Example 9.27

Solve by using the Quadratic Formula: 4 x 2 − 20 x = −25 . 4 x 2 − 20 x = −25 .

Did you recognize that 4 x 2 − 20 x + 25 is a perfect square trinomial. It is equivalent to (2 x − 5) 2 ? If you solve 4 x 2 − 20 x + 25 = 0 by factoring and then using the Square Root Property, do you get the same result?

Try It 9.53

Solve by using the Quadratic Formula: r 2 + 10 r + 25 = 0 . r 2 + 10 r + 25 = 0 .

Try It 9.54

Solve by using the Quadratic Formula: 25 t 2 − 40 t = −16 . 25 t 2 − 40 t = −16 .

Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant .

Discriminant

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.

Using the Discriminant, b 2 − 4 ac , to Determine the Number and Type of Solutions of a Quadratic Equation

For a quadratic equation of the form ax 2 + bx + c = 0, a ≠ 0 , a ≠ 0 ,

• If b 2 − 4 ac > 0, the equation has 2 real solutions.
• if b 2 − 4 ac = 0, the equation has 1 real solution.
• if b 2 − 4 ac < 0, the equation has 2 complex solutions.

Example 9.28

Determine the number of solutions to each quadratic equation.

ⓐ 3 x 2 + 7 x − 9 = 0 3 x 2 + 7 x − 9 = 0 ⓑ 5 n 2 + n + 4 = 0 5 n 2 + n + 4 = 0 ⓒ 9 y 2 − 6 y + 1 = 0 . 9 y 2 − 6 y + 1 = 0 .

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

Since the discriminant is positive, there are 2 real solutions to the equation.

Since the discriminant is negative, there are 2 complex solutions to the equation.

Since the discriminant is 0, there is 1 real solution to the equation.

Try It 9.55

Determine the numberand type of solutions to each quadratic equation.

ⓐ 8 m 2 − 3 m + 6 = 0 8 m 2 − 3 m + 6 = 0 ⓑ 5 z 2 + 6 z − 2 = 0 5 z 2 + 6 z − 2 = 0 ⓒ 9 w 2 + 24 w + 16 = 0 . 9 w 2 + 24 w + 16 = 0 .

Try It 9.56

Determine the number and type of solutions to each quadratic equation.

ⓐ b 2 + 7 b − 13 = 0 b 2 + 7 b − 13 = 0 ⓑ 5 a 2 − 6 a + 10 = 0 5 a 2 − 6 a + 10 = 0 ⓒ 4 r 2 − 20 r + 25 = 0 . 4 r 2 − 20 r + 25 = 0 .

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We summarize the four methods that we have used to solve quadratic equations below.

Methods for Solving Quadratic Equations

• Square Root Property
• Completing the Square

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is a x 2 = k a x 2 = k or a ( x − h ) 2 = k a ( x − h ) 2 = k we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.

Identify the most appropriate method to solve a quadratic equation.

• Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
• Step 2. Try the Square Root Property next. If the equation fits the form a x 2 = k a x 2 = k or a ( x − h ) 2 = k , a ( x − h ) 2 = k , it can easily be solved by using the Square Root Property.
• Step 3. Use the Quadratic Formula . Any other quadratic equation is best solved by using the Quadratic Formula.

The next example uses this strategy to decide how to solve each quadratic equation.

Example 9.29

Identify the most appropriate method to use to solve each quadratic equation.

ⓐ 5 z 2 = 17 5 z 2 = 17 ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0 ⓒ 8 u 2 + 6 u = 11 . 8 u 2 + 6 u = 11 .

ⓐ 5 z 2 = 17 5 z 2 = 17

Since the equation is in the a x 2 = k , a x 2 = k , the most appropriate method is to use the Square Root Property.

ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.

Try It 9.57

ⓐ x 2 + 6 x + 8 = 0 x 2 + 6 x + 8 = 0 ⓑ ( n − 3 ) 2 = 16 ( n − 3 ) 2 = 16 ⓒ 5 p 2 − 6 p = 9 . 5 p 2 − 6 p = 9 .

Try It 9.58

ⓐ 8 a 2 + 3 a − 9 = 0 8 a 2 + 3 a − 9 = 0 ⓑ 4 b 2 + 4 b + 1 = 0 4 b 2 + 4 b + 1 = 0 ⓒ 5 c 2 = 125 . 5 c 2 = 125 .

Access these online resources for additional instruction and practice with using the Quadratic Formula.

• Using the Quadratic Formula
• Solve a Quadratic Equation Using the Quadratic Formula with Complex Solutions
• Discriminant in Quadratic Formula

Section 9.3 Exercises

Practice makes perfect.

In the following exercises, solve by using the Quadratic Formula.

4 m 2 + m − 3 = 0 4 m 2 + m − 3 = 0

4 n 2 − 9 n + 5 = 0 4 n 2 − 9 n + 5 = 0

2 p 2 − 7 p + 3 = 0 2 p 2 − 7 p + 3 = 0

3 q 2 + 8 q − 3 = 0 3 q 2 + 8 q − 3 = 0

p 2 + 7 p + 12 = 0 p 2 + 7 p + 12 = 0

q 2 + 3 q − 18 = 0 q 2 + 3 q − 18 = 0

r 2 − 8 r = 33 r 2 − 8 r = 33

t 2 + 13 t = −40 t 2 + 13 t = −40

3 u 2 + 7 u − 2 = 0 3 u 2 + 7 u − 2 = 0

2 p 2 + 8 p + 5 = 0 2 p 2 + 8 p + 5 = 0

2 a 2 − 6 a + 3 = 0 2 a 2 − 6 a + 3 = 0

5 b 2 + 2 b − 4 = 0 5 b 2 + 2 b − 4 = 0

x 2 + 8 x − 4 = 0 x 2 + 8 x − 4 = 0

y 2 + 4 y − 4 = 0 y 2 + 4 y − 4 = 0

3 y 2 + 5 y − 2 = 0 3 y 2 + 5 y − 2 = 0

6 x 2 + 2 x − 20 = 0 6 x 2 + 2 x − 20 = 0

2 x 2 + 3 x + 3 = 0 2 x 2 + 3 x + 3 = 0

2 x 2 − x + 1 = 0 2 x 2 − x + 1 = 0

8 x 2 − 6 x + 2 = 0 8 x 2 − 6 x + 2 = 0

8 x 2 − 4 x + 1 = 0 8 x 2 − 4 x + 1 = 0

( v + 1 ) ( v − 5 ) − 4 = 0 ( v + 1 ) ( v − 5 ) − 4 = 0

( x + 1 ) ( x − 3 ) = 2 ( x + 1 ) ( x − 3 ) = 2

( y + 4 ) ( y − 7 ) = 18 ( y + 4 ) ( y − 7 ) = 18

( x + 2 ) ( x + 6 ) = 21 ( x + 2 ) ( x + 6 ) = 21

1 3 m 2 + 1 12 m = 1 4 1 3 m 2 + 1 12 m = 1 4

1 3 n 2 + n = − 1 2 1 3 n 2 + n = − 1 2

3 4 b 2 + 1 2 b = 3 8 3 4 b 2 + 1 2 b = 3 8

1 9 c 2 + 2 3 c = 3 1 9 c 2 + 2 3 c = 3

16 c 2 + 24 c + 9 = 0 16 c 2 + 24 c + 9 = 0

25 d 2 − 60 d + 36 = 0 25 d 2 − 60 d + 36 = 0

25 q 2 + 30 q + 9 = 0 25 q 2 + 30 q + 9 = 0

16 y 2 + 8 y + 1 = 0 16 y 2 + 8 y + 1 = 0

Use the Discriminant to Predict the Number of Real Solutions of a Quadratic Equation

In the following exercises, determine the number of real solutions for each quadratic equation.

ⓐ 4 x 2 − 5 x + 16 = 0 4 x 2 − 5 x + 16 = 0 ⓑ 36 y 2 + 36 y + 9 = 0 36 y 2 + 36 y + 9 = 0 ⓒ 6 m 2 + 3 m − 5 = 0 6 m 2 + 3 m − 5 = 0

ⓐ 9 v 2 − 15 v + 25 = 0 9 v 2 − 15 v + 25 = 0 ⓑ 100 w 2 + 60 w + 9 = 0 100 w 2 + 60 w + 9 = 0 ⓒ 5 c 2 + 7 c − 10 = 0 5 c 2 + 7 c − 10 = 0

ⓐ r 2 + 12 r + 36 = 0 r 2 + 12 r + 36 = 0 ⓑ 8 t 2 − 11 t + 5 = 0 8 t 2 − 11 t + 5 = 0 ⓒ 3 v 2 − 5 v − 1 = 0 3 v 2 − 5 v − 1 = 0

ⓐ 25 p 2 + 10 p + 1 = 0 25 p 2 + 10 p + 1 = 0 ⓑ 7 q 2 − 3 q − 6 = 0 7 q 2 − 3 q − 6 = 0 ⓒ 7 y 2 + 2 y + 8 = 0 7 y 2 + 2 y + 8 = 0

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

ⓐ x 2 − 5 x − 24 = 0 x 2 − 5 x − 24 = 0 ⓑ ( y + 5 ) 2 = 12 ( y + 5 ) 2 = 12 ⓒ 14 m 2 + 3 m = 11 14 m 2 + 3 m = 11

ⓐ ( 8 v + 3 ) 2 = 81 ( 8 v + 3 ) 2 = 81 ⓑ w 2 − 9 w − 22 = 0 w 2 − 9 w − 22 = 0 ⓒ 4 n 2 − 10 n = 6 4 n 2 − 10 n = 6

ⓐ 6 a 2 + 14 a = 20 6 a 2 + 14 a = 20 ⓑ ( x − 1 4 ) 2 = 5 16 ( x − 1 4 ) 2 = 5 16 ⓒ y 2 − 2 y = 8 y 2 − 2 y = 8

ⓐ 8 b 2 + 15 b = 4 8 b 2 + 15 b = 4 ⓑ 5 9 v 2 − 2 3 v = 1 5 9 v 2 − 2 3 v = 1 ⓒ ( w + 4 3 ) 2 = 2 9 ( w + 4 3 ) 2 = 2 9

Writing Exercises

Solve the equation x 2 + 10 x = 120 x 2 + 10 x = 120

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?

Solve the equation 12 y 2 + 23 y = 24 12 y 2 + 23 y = 24

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/9-3-solve-quadratic-equations-using-the-quadratic-formula

© Jul 24, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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Perfect Squares and Square Roots Quiz

Complete the square, 8th -  11th  , solving quadratics by completing the squ..., 11th -  12th  , square and cube roots, 7th -  8th  , area of triangles and parallelograms, perfect squares and square roots, 7th -  9th  , scientific notation, 8th -  9th  , square roots.

Solving Quadratic Equations by Completin...

Mathematics.

Solving Quadratic Equations by Completing the Square

30 questions

Introducing new   Paper mode

No student devices needed.   Know more

• 1. Multiple Choice Edit 1.5 minutes 1 pt What do you do to the b value to correctly complete the square? square it divide it by 2 and square the result divide it by 2 and take the square root of the result divide it by 2 only

Complete the square for

x 2 - 12x + ____

Complete the square.

x 2 + 6x + ____

x 2 - 14x + ____

x 2 - 10x + ____

x 2 + 10x - 100

x 2 - 10x + 100

x 2 + 10x - 25

x 2 - 10x + 25

• 6. Multiple Choice Edit 45 seconds 1 pt Find the missing value "c" to create a perfect square trinomial: x 2  + 26x + ___  13 136 169 26
• 7. Multiple Choice Edit 1.5 minutes 1 pt Find the missing value "c" to create a perfect square trinomial: x 2  – 8x  + ___  -4 16 4 8

Given the perfect square trinomial, identify the equivalent expression

x 2 -18x+81

• 9. Multiple Choice Edit 5 minutes 1 pt To solve by completing the square, what needs to be moved in this equation? x 2 = 9 - 4x the -4x the 9 the x 2
• 10. Multiple Choice Edit 30 seconds 1 pt Solve by completing the square: k 2 − 12k + 23 = 0 {6 + √13, 6 - √13} {-6 + √13, -6 - √13} {6 + √59, 6 - √59} {-6 + √59, -6 - √59}
• 11. Multiple Choice Edit 2 minutes 1 pt Solve by completing the square. y 2 + 10y = -9 1 and -12 -1 and -9 1 and -9 1 and -1

Solve by Completing the Square:

2x 2 +4x-5=7

Hint: Move the 5 and you must have a lead number of 1!

-0.4 and -3.4

-2.3 and 4.3

-3.6 and 1.6

no solution

• 13. Multiple Choice Edit 5 minutes 1 pt When factoring x 2  - 4x + 4 = 20, what goes in the blank? (x - __ ) 2 = 20 4 2 8 20
• 15. Multiple Choice Edit 5 minutes 1 pt What should be added to both sides of the following equation to complete the square? x 2  - 12x = 5 -12 6 36 -5
• 16. Multiple Choice Edit 5 minutes 1 pt What is one of the solutions to this equation? (2x - 1) 2 = 49 7 -7 -4 -3

x 2 +14x+49

• 18. Multiple Choice Edit 1.5 minutes 1 pt Solve the equation by completing the square and then finding the roots.  x 2 + 6x - 4 = 36 x = -7, 7 x = 4, -10 x = -4 +- √7 x = 10, -4
• 19. Multiple Choice Edit 30 seconds 1 pt Solve the following equation by completing the square: n 2 - 2n - 3 = 0 {3 and-1} {4 and -4} {5 and -3} {8 and -7}
• 20. Multiple Choice Edit 3 minutes 1 pt Solve by completing the square: k 2 − 12k + 23 = 0 {6 + √13, 6 - √13} {-6 + √13, -6 - √13} {6 + √59, 6 - √59} {-6 + √59, -6 - √59}
• 21. Multiple Choice Edit 30 seconds 1 pt Solve by completing the square. y 2 + 10y = -9 1 and -12 -1 and -9 1 and -9 1 and -1
• 22. Multiple Choice Edit 30 seconds 1 pt What is the first step to solving THIS equation by completing the square? a 2 + 10a + 21 = 0 Set the equation equal to zero Divide 10 by 2 and add the result to both sides Add a 2 and 10a together Subtract the 21
• 23. Multiple Choice Edit 1.5 minutes 1 pt When factoring x 2  - 4x + 4 = 20, what goes in the blank? (x - __ ) 2 = 20 4 2 8 20
• 24. Multiple Choice Edit 30 seconds 1 pt What is one of the solutions to this equation? (2x - 1) 2 = 49 7 -7 -4 -3
• 25. Multiple Choice Edit 3 minutes 1 pt Solve the following equation by completing the square: n 2 - 2n - 3 = 0 {3 and-1} {4 and -4} {5 and -3} {8 and -7}
• 26. Multiple Choice Edit 15 minutes 1 pt What is the first step to solving THIS equation by completing the square? a 2 + 10a + 21 = 0 Set the equation equal to zero Divide 10 by 2 and add the result to both sides Add a 2 and 10a together Subtract the 21
• 27. Multiple Choice Edit 15 minutes 1 pt What is one of the solutions to this equation? (2x - 1) 2 = 49 7 -7 -4 -3

Practice #2: a 2 - 6a + 12 = 0

x = 3 and x = 1

x = -3 and x = 4

x = 12 and x = 2

• 29. Multiple Choice Edit 30 seconds 1 pt Solve the following equation by completing the square: n 2 = 18n + 40 {11 and -11} {16 and 2} {20 and -2} {10 and -4}

x 2 −16x−33=3

x= 18, x= −2

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