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Completing the Square

Create perfect square trinomials using the additive property of equality

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Practice Completing the Square
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Completing the Square When the Leading Coefficient Doesn't Equal 1

The area of another parallelogram is given by the equation 3x^2 + 9x - 5 = 0 , where x is the length of the base. What is the length of this base?

Guidance

When there is a number in front of @$x^2@$ , it will make completing the square a little more complicated. See how the steps change in Example A.

Example A

Determine the number c that completes the square of @$2x^2 - 8x + c @$ .

Solution: In the previous concept, we just added @$\left(\frac{b}{2}\right)^2@$ , but that was when @$a=1@$ . Now that @$a \neq 1@$ , we have to take the value of a into consideration. Let's pull out the GCF of 2 and 8 first.

@$2 \left(x^2 - 4x \right)@$

Now, there is no number in front of @$x^2@$ .

@$\left(\frac{b}{2}\right)^2 = \left(\frac{4}{2}\right)^2 = 4@$ .

Add this number inside the parenthesis and distribute the 2.

@$2 \left(x^2 - 4x +4 \right)=2x^2-4x+8@$

So, @$c=8@$ .

Example B

Solve @$3x^2-9x+11=0@$

Solution:

1. Write the polynomial so that @$x^2@$ and @$x@$ are on the left side of the equation and the constants on the right.

@$$3x^2-9x=-11@$$

2. Pull out @$a@$ from everything on the left side. Even if @$b@$ is not divisible by @$a@$ , the coefficient of @$x^2@$ needs to be 1 in order to complete the square.

@$$3(x^2-3x+\underline{\;\;\;\;\;\;})=-11@$$

3. Now, complete the square. Determine what number would make a perfect square trinomial.

To do this, divide the @$x-@$ term by 2 and square that number, or @$\left(\frac{b}{2}\right)^2@$ .

@$$\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}@$$

4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add @${\color{red}a\cdot\left(\frac{b}{2}\right)^2}@$ to keep the equation balanced.

@$$3\left(x^2-3x {\color{red}+\frac{9}{4}}\right)=-11 {\color{red}+\frac{27}{4}}@$$

5. Factor the left side and simplify the right.

@$$3\left(x-\frac{3}{2}\right)^2=-\frac{17}{4}@$$

6. Solve by using square roots.

@$$\left(x-\frac{3}{2}\right)^2 &=-\frac{17}{12}\\ x-\frac{3}{2} &=\pm \frac{i\sqrt{17}}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\ x &=\frac{3}{2}\pm \frac{\sqrt{51}}{6}i@$$

Be careful with the addition of step 2 and the changes made to step 4. A very common mistake is to add @$\left(\frac{b}{2}\right)^2@$ to both sides, without multiplying by @$a@$ for the right side.

Example C

Solve @$4x^2+7x-18=0@$ .

Solution: Let’s follow the steps from Example B.

1. Write the polynomial so that @$x^2@$ and @$x@$ are on the left side of the equation and the constants on the right.

@$$4x^2-7x=18@$$

2. Pull out @$a@$ from everything on the left side.

@$$4\left(x^2+\frac{7}{4}x+\underline{\;\;\;\;\;\;}\right)=18@$$

3. Now, complete the square. Find @$\left(\frac{b}{2}\right)^2@$ .

@$$\left(\frac{b}{2}\right)^2=\left(\frac{7}{8}\right)^2=\frac{49}{64}@$$

4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add @${\color{red}a\cdot\left(\frac{b}{2}\right)^2}@$ to keep the equation balanced.

@$$4\left(x^2+\frac{7}{4}x{\color{red}+\frac{49}{64}}\right)=18{\color{red}+\frac{49}{16}}@$$

5. Factor the left side and simplify the right.

@$$4\left(x+\frac{7}{8}\right)^2=\frac{337}{16}@$$

6. Solve by using square roots.

@$$\left(x+\frac{7}{8}\right)^2 &=\frac{337}{64}\\ x+\frac{7}{8} &=\pm \frac{\sqrt{337}}{8}\\ x &=-\frac{7}{8} \pm \frac{\sqrt{337}}{8}@$$

Intro Problem Revisit We can't factor @$3x^2 + 9x - 5 = 0@$ , so let's follow the step-by-step process we learned in this lesson.

1. Write the polynomial so that @$x^2@$ and @$x@$ are on the left side of the equation and the constants on the right.

@$$3x^2 + 9x = 5@$$

2. Pull out @$a@$ from everything on the left side.

@$$3\left(x^2 + 3x+\underline{\;\;\;\;\;\;}\right) = 5@$$

3. Now, complete the square. Find @$\left(\frac{b}{2}\right)^2@$ .

@$$\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}@$$

4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add @${\color{red}a\cdot\left(\frac{b}{2}\right)^2}@$ to keep the equation balanced.

@$$3\left(x^2+ 3x{\color{red}+\frac{9}{4}}\right)=5{\color{red}+\frac{27}{4}}@$$

5. Factor the left side and simplify the right.

@$$3\left(x+\frac{3}{2}\right)^2=\frac{47}{4}@$$

6. Solve by using square roots.

@$$\left(x+\frac{3}{2}\right)^2 &=\frac{47}{12}\\ x+\frac{3}{2} &=\pm \frac{\sqrt{47}}{\sqrt{12}}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt{47}}}{2\sqrt{3}}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt{141}}}{6} @$$

However, because x is the length of the parallelogram's base, it must have a positive value. Only @$x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6}@$ results in a positive value, so the length of the base is @$x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6}@$ .

Guided Practice

Solve the following quadratic equations by completing the square.

1. @$5x^2+29x-6=0@$

2. @$8x^2-32x+4=0@$

Answers

Use the steps from the examples above to solve for @$x@$ .

1.

@$$5x^2+29x-6 &=0\\ 5\left(x^2+\frac{29}{5}x\right) &=6\\ 5\left(x^2+\frac{29}{5}x+\frac{841}{100}\right) &=6+\frac{841}{20}\\ 5\left(x+\frac{29}{10}\right)^2 &=\frac{961}{20}\\ \left(x+\frac{29}{10}\right)^2 &=\frac{961}{100}\\ x+\frac{29}{10} &=\pm \frac{31}{10}\\ x &=-\frac{29}{10} \pm \frac{31}{10}\\ x &=-6, \frac{1}{5}@$$

2.

@$$8x^2-32x+4 &=0\\ 8(x^2-4x) &=-4\\ 8(x^2-4x+4) &=-4+32\\ 8(x-2)^2 &=28\\ (x-2)^2 &=\frac{7}{2}\\ x-2 &=\pm \frac{\sqrt{7}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\ x &=2\pm \frac{\sqrt{14}}{2}@$$

Explore More

Solve the quadratic equations by completing the square.

  1. @$6x^2-12x-7=0@$
  2. @$-4x^2+24x-100=0@$
  3. @$5x^2-30x+55=0@$
  4. @$2x^2-x-6=0@$
  5. @$\frac{1}{2}x^2+7x+8=0@$
  6. @$-3x^2+4x+15=0@$

Solve the following equations by factoring, using square roots, or completing the square.

  1. @$4x^2-4x-8=0@$
  2. @$2x^2+9x+7=0@$
  3. @$-5(x+4)^2-19=26@$
  4. @$3x^2+30x-5=0@$
  5. @$9x^2-15x-6=0@$
  6. @$10x^2+40x+88=0@$

Problems 13-15 build off of each other.

  1. Challenge Complete the square for @$ax^2+bx+c=0@$ . Follow the steps from Examples A and B. Your final answer should be in terms of @$a, b,@$ and @$c@$ .
  2. For the equation @$8x^2+6x-5=0@$ , use the formula you found in #13 to solve for @$x@$ .
  3. Is the equation in #14 factorable? If so, factor and solve it.
  4. Error Analysis Examine the worked out problem below.

@$$4x^2-48x+11&=0\\ 4(x^2-12x+\underline{\;\;\;\;\;\;}) &=-11\\ 4(x^2-12x+36) &=-11+36\\ 4(x-6)^2 &=25\\ (x-6)^2 &=\frac{25}{4}\\ x-6 &=\pm \frac{5}{2}\\ x &=6\pm \frac{5}{2} \rightarrow \frac{17}{2},\frac{7}{2}@$$

Plug the answers into the original equation to see if they work. If not, find the error and correct it.

Vocabulary

Perfect Square Trinomial

Perfect Square Trinomial

A perfect square trinomial is a quadratic expression of the form a^2+2ab+b^2 (which can be rewritten as (a+b)^2) or a^2-2ab+b^2 (which can be rewritten as (a-b)^2).
Square Root

Square Root

The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9.

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