Question

Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x+27)^2/−6 = −3 .(1 point)

Answers

Answer
there's only one answer?
Answered by GPT-4o mini
To solve the quadratic equation \(\frac{(x+27)^2}{-6} = -3\), start by isolating the squared term.

1. Multiply both sides of the equation by \(-6\) to eliminate the fraction:

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

This simplifies to:

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

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

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

3. Simplify \(\sqrt{18}\):

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

Now we have:

\[
x + 27 = \pm 3\sqrt{2}
\]

4. Solve for \(x\) by subtracting 27 from both sides:

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

This gives us two solutions:

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

5. Now, calculate the numerical values of both solutions:

- Calculate \(3\sqrt{2}\):

\[
3\sqrt{2} \approx 3 \times 1.414 \approx 4.242
\]

- Therefore, the two solutions are:

\[
x \approx -27 + 4.242 \approx -22.758
\]
\[
x \approx -27 - 4.242 \approx -31.242
\]

6. Rounding to the nearest hundredth:

- For \(x \approx -22.758\):

\[
x \approx -22.76
\]

- For \(x \approx -31.242\):

\[
x \approx -31.24
\]

Thus, the final answers are:

\[
x \approx -22.76 \quad \text{and} \quad x \approx -31.24
\]
Answered by GPT-4o mini
The quadratic equation we solved indeed gives us two solutions, but let me clarify what that means in terms of the context of the problem.

From the equation:

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

Both of these yield valid answers, and they represent two distinct values of \(x\).

### Summary of Solutions:

1. **First Solution**:
\[
x = -27 + 3\sqrt{2} \approx -22.76
\]

2. **Second Solution**:
\[
x = -27 - 3\sqrt{2} \approx -31.24
\]

### Conclusion:

Both solutions are valid, leading us to conclude that there are indeed **two answers**:

- \(x \approx -22.76\)
- \(x \approx -31.24\)

If you were inquiring about a single specific value, sometimes in a context, you might be asked for the principal (or first) solution, or maybe only one is of interest based on additional context. However, both solutions are equally valid for the equation as presented.

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