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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 3x2+9x5=0, 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 x2, it will make completing the square a little more complicated. 

Let's determine the number c that completes the square of 2x28x+c.

Previously, we just added (b2)2, but that was when a=1. Now that a1, we have to take the value of a into consideration. Let's pull out the GCF of 2 and 8 first.

2(x24x)

Now, there is no number in front of x2.

(b2)2=(42)2=4.

Add this number inside the parenthesis and distribute the 2.

2(x24x+4)=2x24x+8

So, c=8.

Now, let's solve the following problems by completing the square.

  1. Solve 3x29x+11=0.

Step 1: Write the polynomial so that x2 and x are on the left side of the equation and the constants on the right.

3x29x=11

Step 2: Pull out a from everything on the left side. Even if b is not divisible by a, the coefficient of x2 needs to be 1 in order to complete the square.

3(x23x+)=11

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

To do this, divide the xterm by 2 and square that number, or (b2)2.

(b2)2=(32)2=94

Step 4: Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add a(b2)2 to keep the equation balanced.

3(x23x+94)=11+274

Step 5: Factor the left side and simplify the right.

3(x32)2=174

Step 6: Solve by using square roots.

(x32)2x32x=1712=±i172333=32±516i

Be careful with the addition of Step 2 and the changes made to Step 4. A very common mistake is to add (b2)2 to both sides, without multiplying by a for the right side.

  1. Solve 4x2+7x18=0.

Let’s follow the steps from problem #1 above.

Step 1: Write the polynomial so that x2 and x are on the left side of the equation and the constants on the right.

4x27x=18

Step 2: Pull out a from everything on the left side.

4(x2+74x+)=18

Step 3: Now, complete the square. Find (b2)2.

(b2)2=(78)2=4964

Step 4: Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add a(b2)2 to keep the equation balanced.

4(x2+74x+4964)=18+4916

Step 5: Factor the left side and simplify the right.

4(x+78)2=33716

Step 6: Solve by using square roots.

(x+78)2x+78x=33764=±3378=78±3378

Examples

Example 1

Earlier, you were asked to find the length of the base of the parallelogram. 

We can't factor 3x2+9x5=0, so let's follow the step-by-step process we learned in this lesson.

Step 1: Write the polynomial so that x2 and x are on the left side of the equation and the constants on the right.

3x2+9x=5

Step 2: Pull out a from everything on the left side.

3(x2+3x+)=5

Step 3: Now, complete the square. Find (b2)2.

(b2)2=(32)2=94

Step 4: Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add a(b2)2 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*}3(x2+3x+94)=5+274

Step 5: Factor the left side and simplify the right.

\begin{align*}3\left(x+\frac{3}{2}\right)^2=\frac{47}{4}\end{align*}3(x+32)2=474

Step 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*}(x+32)2x+32xx=4712=±4712=32±4723=32±1416

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*}x=32+1416 results in a positive value, so the length of the base is \begin{align*}x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6}\end{align*}x=32+1416.

Example 2

Solve the following quadratic equation by completing the square: \begin{align*}5x^2+29x-6=0\end{align*}5x2+29x6=0.

\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*}5x2+29x65(x2+295x)5(x2+295x+841100)5(x+2910)2(x+2910)2x+2910xx=0=6=6+84120=96120=961100=±3110=2910±3110=6,15

Example 3

Solve the following quadratic by completing the square: \begin{align*}8x^2-32x+4=0\end{align*}8x232x+4=0.

\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*}8x232x+48(x24x)8(x24x+4)8(x2)2(x2)2x2x=0=4=4+32=28=72=±7222=2±142

Review

Solve the quadratic equations by completing the square.

  1. \begin{align*}6x^2-12x-7=0\end{align*}6x212x7=0
  2. \begin{align*}-4x^2+24x-100=0\end{align*}4x2+24x100=0
  3. \begin{align*}5x^2-30x+55=0\end{align*}5x230x+55=0
  4. \begin{align*}2x^2-x-6=0\end{align*}2x2x6=0
  5. \begin{align*}\frac{1}{2}x^2+7x+8=0\end{align*}12x2+7x+8=0
  6. \begin{align*}-3x^2+4x+15=0\end{align*}3x2+4x+15=0

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

  1. \begin{align*}4x^2-4x-8=0\end{align*}4x24x8=0
  2. \begin{align*}2x^2+9x+7=0\end{align*}2x2+9x+7=0
  3. \begin{align*}-5(x+4)^2-19=26\end{align*}5(x+4)219=26
  4. \begin{align*}3x^2+30x-5=0\end{align*}3x2+30x5=0
  5. \begin{align*}9x^2-15x-6=0\end{align*}9x215x6=0
  6. \begin{align*}10x^2+40x+88=0\end{align*}10x2+40x+88=0

Problems 13-15 build off of each other.

  1. Challenge Complete the square for \begin{align*}ax^2+bx+c=0\end{align*}ax2+bx+c=0. Follow the steps outlined in this lesson. Your final answer should be in terms of \begin{align*}a, b,\end{align*}a,b, and \begin{align*}c\end{align*}c.
  2. For the equation \begin{align*}8x^2+6x-5=0\end{align*}8x2+6x5=0, use the formula you found in #13 to solve for \begin{align*}x\end{align*}x.
  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*}4x248x+114(x212x+)4(x212x+36)4(x6)2(x6)2x6x=0=11=11+36=25=254=±52=6±52172,72

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. 

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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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