Eliza solved a quadratic equation. Her work is shown below, with Step 2 missing.

What could Eliza have written as the result from Step 2?

[tex]\[
\begin{array}{l}
\frac{1}{5}(x+3)^2 - 2 = 18 \\
\frac{1}{5}(x+3)^2 = 20 \quad \text{Step 1} \\
(x+3)^2 = 100 \quad \text{Step 2} \\
x+3 = \pm 10 \quad \text{Step 3} \\
x = -13 \text{ or } x = 7 \quad \text{Step 4} \\
\end{array}
\][/tex]



Answer :

Let's go through the steps Eliza might have taken to solve the quadratic equation in detail.

We start with the original equation:
[tex]\[ \frac{1}{5}(x+3)^2 - 2 = 18 \][/tex]

For Step 1, we need to isolate the square term. First, we add 2 to both sides:
[tex]\[ \frac{1}{5}(x+3)^2 - 2 + 2 = 18 + 2 \][/tex]
[tex]\[ \frac{1}{5}(x+3)^2 = 20 \][/tex]

Next, we multiply both sides of the equation by 5 to eliminate the fraction:
[tex]\[ 5 \cdot \frac{1}{5}(x+3)^2 = 5 \cdot 20 \][/tex]
[tex]\[ (x+3)^2 = 100 \][/tex]

This brings us to Step 2. Therefore, Eliza should have written:
[tex]\[ (x+3)^2 = 100 \][/tex]

Step 3 involves solving for [tex]\( x+3 \)[/tex] by taking the square root of both sides:
[tex]\[ x+3 = \pm \sqrt{100} \][/tex]
[tex]\[ x+3 = \pm 10 \][/tex]

Step 4 is to solve for [tex]\( x \)[/tex]:
[tex]\[ x+3 = 10 \quad \Rightarrow \quad x = 10 - 3 = 7 \][/tex]
[tex]\[ x+3 = -10 \quad \Rightarrow \quad x = -10 - 3 = -13 \][/tex]

Thus, the solutions are:
[tex]\[ x = -13 \text{ or } x = 7 \][/tex]

So, in summary, the result from Step 2 should be:
[tex]\[ (x+3)^2 = 100 \][/tex]