# 5.12: Completing the Square When the Leading Coefficient Doesn't Equal 1

Difficulty Level: Advanced Created by: CK-12
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Practice Completing the Square

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

### Guidance

When there is a number in front of \begin{align*}x^2\end{align*}, 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 \begin{align*}2x^2 - 8x + c \end{align*}.

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

\begin{align*}2 \left(x^2 - 4x \right)\end{align*}

Now, there is no number in front of \begin{align*}x^2\end{align*}.

\begin{align*}\left(\frac{b}{2}\right)^2 = \left(\frac{4}{2}\right)^2 = 4\end{align*}.

Add this number inside the parenthesis and distribute the 2.

\begin{align*}2 \left(x^2 - 4x +4 \right)=2x^2-4x+8\end{align*}

So, \begin{align*}c=8\end{align*}.

#### Example B

Solve \begin{align*}3x^2-9x+11=0\end{align*}

Solution:

1. Write the polynomial so that \begin{align*}x^2\end{align*} and \begin{align*}x\end{align*} are on the left side of the equation and the constants on the right.

\begin{align*}3x^2-9x=-11\end{align*}

2. Pull out \begin{align*}a\end{align*} from everything on the left side. Even if \begin{align*}b\end{align*} is not divisible by \begin{align*}a\end{align*}, the coefficient of \begin{align*}x^2\end{align*} needs to be 1 in order to complete the square.

\begin{align*}3(x^2-3x+\underline{\;\;\;\;\;\;})=-11\end{align*}

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

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

\begin{align*}\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}\end{align*}

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

\begin{align*}3\left(x^2-3x {\color{red}+\frac{9}{4}}\right)=-11 {\color{red}+\frac{27}{4}}\end{align*}

5. Factor the left side and simplify the right.

\begin{align*}3\left(x-\frac{3}{2}\right)^2=-\frac{17}{4}\end{align*}

6. Solve by using square roots.

\begin{align*}\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\end{align*}

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

#### Example C

Solve \begin{align*}4x^2+7x-18=0\end{align*}.

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

1. Write the polynomial so that \begin{align*}x^2\end{align*} and \begin{align*}x\end{align*} are on the left side of the equation and the constants on the right.

\begin{align*}4x^2-7x=18\end{align*}

2. Pull out \begin{align*}a\end{align*} from everything on the left side.

\begin{align*}4\left(x^2+\frac{7}{4}x+\underline{\;\;\;\;\;\;}\right)=18\end{align*}

3. Now, complete the square. Find \begin{align*}\left(\frac{b}{2}\right)^2\end{align*}.

\begin{align*}\left(\frac{b}{2}\right)^2=\left(\frac{7}{8}\right)^2=\frac{49}{64}\end{align*}

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

\begin{align*}4\left(x^2+\frac{7}{4}x{\color{red}+\frac{49}{64}}\right)=18{\color{red}+\frac{49}{16}}\end{align*}

5. Factor the left side and simplify the right.

\begin{align*}4\left(x+\frac{7}{8}\right)^2=\frac{337}{16}\end{align*}

6. Solve by using square roots.

\begin{align*}\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}\end{align*}

Intro Problem Revisit We can't factor \begin{align*}3x^2 + 9x - 5 = 0\end{align*}, so let's follow the step-by-step process we learned in this lesson.

1. Write the polynomial so that \begin{align*}x^2\end{align*} and \begin{align*}x\end{align*} are on the left side of the equation and the constants on the right.

\begin{align*}3x^2 + 9x = 5\end{align*}

2. Pull out \begin{align*}a\end{align*} from everything on the left side.

\begin{align*}3\left(x^2 + 3x+\underline{\;\;\;\;\;\;}\right) = 5\end{align*}

3. Now, complete the square. Find \begin{align*}\left(\frac{b}{2}\right)^2\end{align*}.

\begin{align*}\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}\end{align*}

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

\begin{align*}3\left(x^2+ 3x{\color{red}+\frac{9}{4}}\right)=5{\color{red}+\frac{27}{4}}\end{align*}

5. Factor the left side and simplify the right.

\begin{align*}3\left(x+\frac{3}{2}\right)^2=\frac{47}{4}\end{align*}

6. Solve by using square roots.

\begin{align*}\left(x+\frac{3}{2}\right)^2 &=\frac{47}{12}\\ x+\frac{3}{2} &=\pm \frac{\sqrt{47}}{\sqrt12}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt47}}{2\sqrt3}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt141}}{6} \end{align*}

However, because x is the length of the parallelogram's base, it must have a positive value. Only \begin{align*}x =-\frac{3}{2} + \frac{{\sqrt141}}{6}\end{align*} results in a positive value, so the length of the base is \begin{align*}x =-\frac{3}{2} + \frac{{\sqrt141}}{6}\end{align*}.

### Guided Practice

Solve the following quadratic equations by completing the square.

1. \begin{align*}5x^2+29x-6=0\end{align*}

2. \begin{align*}8x^2-32x+4=0\end{align*}

Use the steps from the examples above to solve for \begin{align*}x\end{align*}.

1.

\begin{align*}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}\end{align*}

2.

\begin{align*}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}\end{align*}

### Practice

Solve the quadratic equations by completing the square.

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

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

1. \begin{align*}4x^2-4x-8=0\end{align*}
2. \begin{align*}2x^2+9x+7=0\end{align*}
3. \begin{align*}-5(x+4)^2-19=26\end{align*}
4. \begin{align*}3x^2+30x-5=0\end{align*}
5. \begin{align*}9x^2-15x-6=0\end{align*}
6. \begin{align*}10x^2+40x+88=0\end{align*}

Problems 13-15 build off of each other.

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

\begin{align*}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}\end{align*}

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

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### Vocabulary Language: English

TermDefinition
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 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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