# Lesson 16

### 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 $$n$$?

1. $$x^2+10x+25$$
2. $$x^2 + 10x + 29$$
3. $$x^2-6x+8$$
4. $$x^2-6x+9$$

### 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.

1. $$x^2 = 100$$
2. $$x^2 = 38$$
3. $$x^2 - 10x + 25 = 0$$
4. $$x^2 + 14x + 40 = 0$$
5. $$x^2 + 14x + 39 = 0$$
6. $$3x^2 - 5x - 11 = 0$$

### Summary

$$\displaystyle x^2 - 5x = 25$$

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.