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# Completing the Square

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

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.