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Completing the Square

Create perfect square trinomials using the additive property of equality

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

The area of a 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?

Completing the Square

When there is a number in front of \begin{align*}x^2\end{align*}, it will make completing the square a little more complicated. 

Solve the following problems by completing the square

Determine the number c that completes the square of \begin{align*}2x^2 - 8x + c \end{align*}.

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 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*}.

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

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.

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

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*}

Examples

Example 1

Earlier, you were asked what is the length of the base. 

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}}{\sqrt{12}}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt{47}}}{2\sqrt{3}}\\ x &=-\frac{3}{2} \pm \frac{{\sqrt{141}}}{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{{\sqrt{141}}}{6}\end{align*} results in a positive value, so the length of the base is \begin{align*}x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6}\end{align*}.

Solve the following quadratic equations by completing the square.

Example 2

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

\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*}

Example 3

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

\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*}

Review

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.

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.12. 

Vocabulary

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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