# Lesson 16

Solving Quadratics

- Let’s solve quadratic equations.

### 16.1: Find the Perfect Squares

The expression \(x^2+8x+16\) is equivalent to \((x+4)^2\). Which expressions are equivalent to \((x+n)^2\) for some number \(n\)?

- \(x^2+10x+25\)
- \(x^2 + 10x + 29\)
- \(x^2-6x+8\)
- \(x^2-6x+9\)

### 16.2: Different Ways to Solve It

Elena and Han solved the equation \(x^2-6x+7=0\) in different ways.

Elena said, “First I added 2 to each side:

\(\displaystyle x^2 - 6x + 7 + 2 = 2\)

So that tells me:

\(\displaystyle (x - 3)^2 = 2\)

I can find the square roots of both sides:

\(\displaystyle x - 3 = \pm \sqrt{2}\)

Which is the same as:

\(\displaystyle x = 3 \pm \sqrt{2}\)

So the two solutions are \(x=3+\sqrt{2}\) and \(x=3-\sqrt{2}\).”

Han said, “I used the quadratic formula:

\(\displaystyle x = \dfrac{\text- b \pm \sqrt{b^2 - 4 \boldcdot a \boldcdot c}}{2 \boldcdot a} \)

Since \(x^2 - 6x + 7 = 0\), that means \(a = 1\), \(b = \text- 6\), and \(c = 7\). I know:

\(\displaystyle x = \dfrac{6 \pm \sqrt{36 - 4 \boldcdot 1 \boldcdot 7}}{2 \boldcdot 1} \)

or

\(\displaystyle x = \dfrac{6 \pm \sqrt{8}}{2} \)

So:

\(\displaystyle x = 3 \pm \frac{\sqrt{8}}{2}\)

I think the solutions are \(x = 3 + \frac{\sqrt{8}}{2}\) and \(x = 3 - \frac{\sqrt{8}}{2}\).”

Do you agree with either of them? Explain your reasoning.

Under what circumstances would solving an equation of the form \(x^2+bx+c=0\) lead to a solution that doesn’t involve fractions?

### 16.3: Solve These Ones

Solve each quadratic equation with the method of your choice. Be prepared to compare your approach with a partner‘s.

- \(x^2 = 100\)
- \(x^2 = 38\)
- \(x^2 - 10x + 25 = 0\)
- \(x^2 + 14x + 40 = 0\)
- \(x^2 + 14x + 39 = 0\)
- \(3x^2 - 5x - 11 = 0\)

### Summary

Consider the quadratic equation:

\(\displaystyle x^2 - 5x = 25\)

It is often most efficient to solve equations like this by completing the square. To complete the square, note that the perfect square \((x+n)^2\) is equal to \(x^2+(2n)x +n^2\). Compare the coefficients of \(x\) in \(x^2+(2n)x +n^2\) to our expression \(x^2-5x\) to see that we want \(2n=\text-5\), or just \(n=\text-\frac{5}{2}\). This means the perfect square \(\left(x-\frac{5}{2}\right)^2\) is equal to \(x^2 -5x + \frac{25}{4}\), so adding \(\frac{25}{4}\) to each side of our equation will give us a perfect square.

\(\displaystyle \begin{align} x^2 - 5x &= 25 \\ x^2- 5x + \frac{25}{4} &= 25 + \frac{25}{4} \\ \left(x-\frac52\right)^2 &= \frac{100}{4} +\frac{25}{4} \\ \left(x-\frac52\right)^2 &= \frac{125}{4} \end{align}\)

The two numbers that square to make \(\frac{125}{4}\) are \(\frac{\sqrt{125}}{2}\) and \(\text-\frac{\sqrt{125}}{2}\), so:

\(\displaystyle x-\frac52 = \pm \frac{\sqrt{125}}{2} \)

which means the two solutions are:

\(\displaystyle x = \frac52 \pm \frac{\sqrt{125}}{2} \)

Other times, it is most efficient to use the quadratic formula. Look at the quadratic equation:

\(\displaystyle 3x^2-2x = 0.8 \)

We could divide each side by 3 and then complete the square like before, but the equation would get even messier and the chance of making a mistake might be higher. With messier equations like this, it is often most efficient to use the quadratic formula:

\(\displaystyle x = {\text-b \pm \sqrt{b^2-4ac} \over 2a} \)

To use this formula, we first need to put the equation in standard form and identify \(a\), \(b\), and \(c\). Rearranging, we get:

\(\displaystyle 3x^2 - 2x -0.8 =0 \)

so \(a=3\), \(b=\text-2\), and \(c=\text-0.8\). We have to be careful to pay attention to the negative signs. Using the quadratic formula, we get:

\(\displaystyle x = {\text-(\text-2) \pm \sqrt{(\text-2)^2-4(3)(\text-0.8)} \over 2(3)} \)

\(\displaystyle x = {2 \pm \sqrt{4+(12)(0.8)} \over 6} \)

Evaluating these solutions with a calculator gives decimal approximations -0.281 and 0.948.