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

Answer 1

To solve the quadratic equation \(\frac{(x + 27)^2}{-6} = -3\), we can first eliminate the fraction by multiplying both sides by \(-6\):

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

This simplifies to:

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

Next, we take the square root of both sides:

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

The square root of 18 can be simplified:

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

So we have two cases:

1. \(x + 27 = 3\sqrt{2}\)
2. \(x + 27 = -3\sqrt{2}\)

Now, we can solve for \(x\) in each case.

Case 1: \[ x + 27 = 3\sqrt{2} \] \[ x = 3\sqrt{2} - 27 \]

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

Now we can approximate the values. First, calculate \(3\sqrt{2}\):

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

Now substituting in the equations:

For Case 1:

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

For Case 2:

Now calculate \(-3\sqrt{2}\):

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

So,

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

Now, we round both answers to the nearest hundredth:

1. Case 1: \(x \approx -22.76\)
2. Case 2: \(x \approx -31.24\)

Thus, the final solutions to the quadratic equation are:

\[ \boxed{-22.76} \quad \text{and} \quad \boxed{-31.24} \]

Answer 2

do I put both the answers

Answer 3

Yes, you should include both answers. The quadratic equation \(\frac{(x + 27)^2}{-6} = -3\) has two solutions for \(x\) due to the square root operation. Thus, you should present both solutions, which are:

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

So you can write your final answer as:

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

Both values are valid solutions to the equation!