Question

Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x+27)2/−6=−3
. (star fraction (x + 27)^2 over -6 end fraction equals -3)

Answer 1

To solve the quadratic equation \(\frac{(x + 27)^2}{-6} = -3\), we'll first isolate the expression on one side.

1. Start by multiplying both sides by \(-6\) to eliminate the fraction:

\[
(x + 27)^2 = -3 \times -6
\]

\[
(x + 27)^2 = 18
\]

2. Next, take the square root of both sides:

\[
x + 27 = \pm \sqrt{18}
\]

We can simplify \(\sqrt{18}\):

\[
\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}
\]

3. Now we have two possible equations:

\[
x + 27 = 3\sqrt{2} \quad \text{or} \quad x + 27 = -3\sqrt{2}
\]

4. Solve for \(x\) in each case:

**First case:**

\[
x = 3\sqrt{2} - 27
\]

**Second case:**

\[
x = -3\sqrt{2} - 27
\]

5. Now we will calculate the numerical solutions:

\[
3\sqrt{2} \approx 3 \cdot 1.414 \approx 4.242
\]
Thus,

\[
x \approx 4.242 - 27 \approx -22.758 \quad \text{(first solution)}
\]

And for the second solution:

\[
-3\sqrt{2} \approx -4.242
\]
So,

\[
x \approx -4.242 - 27 \approx -31.242 \quad \text{(second solution)}
\]

6. Round both solutions to the nearest hundredth:

- First solution: \(x \approx -22.76\)
- Second solution: \(x \approx -31.24\)

Therefore, the solutions to the quadratic equation are approximately:

\[
x \approx -22.76 \quad \text{and} \quad x \approx -31.24
\]