Question

(x-6)^2=51

Answer 1

(x+16)^2/9=7

Answer 2

To solve the equation \((x-6)^2 = 51\), we can start by taking the square root of both sides. First, we rewrite the equation:

\[
x - 6 = \pm \sqrt{51}
\]

Next, we solve for \(x\) by isolating it:

1. For the positive square root:
\[
x - 6 = \sqrt{51}
\]
\[
x = 6 + \sqrt{51}
\]

2. For the negative square root:
\[
x - 6 = -\sqrt{51}
\]
\[
x = 6 - \sqrt{51}
\]

Thus, the solutions are:

\[
x = 6 + \sqrt{51} \quad \text{and} \quad x = 6 - \sqrt{51}
\]

If you'd like to approximate the decimal values, we can calculate \(\sqrt{51} \approx 7.14\):

1. \(x \approx 6 + 7.14 = 13.14\)
2. \(x \approx 6 - 7.14 = -1.14\)

Therefore, the approximate solutions are:

\[
x \approx 13.14 \quad \text{and} \quad x \approx -1.14
\]

Answer 3

To solve the equation

\[
\frac{(x + 16)^2}{9} = 7,
\]

we'll first eliminate the fraction by multiplying both sides by 9:

\[
(x + 16)^2 = 63.
\]

Next, we'll take the square root of both sides:

\[
x + 16 = \pm \sqrt{63}.
\]

Now, we simplify \(\sqrt{63}\):

\[
\sqrt{63} = \sqrt{9 \cdot 7} = 3\sqrt{7}.
\]

So, we have:

\[
x + 16 = \pm 3\sqrt{7}.
\]

Next, we solve for \(x\):

1. For the positive case:

\[
x + 16 = 3\sqrt{7} \implies x = 3\sqrt{7} - 16.
\]

2. For the negative case:

\[
x + 16 = -3\sqrt{7} \implies x = -3\sqrt{7} - 16.
\]

Thus, the solutions for \(x\) are:

\[
x = 3\sqrt{7} - 16 \quad \text{and} \quad x = -3\sqrt{7} - 16.
\]

If you would like approximate decimal values, we calculate \(\sqrt{7} \approx 2.64575\):

1. Calculating \(3\sqrt{7} \approx 3 \times 2.64575 \approx 7.93725\):
- \(x \approx 7.93725 - 16 \approx -8.06275\).

2. For the negative case:
- \(x \approx -7.93725 - 16 \approx -23.93725\).

Thus, the approximate solutions are:

\[
x \approx -8.06 \quad \text{and} \quad x \approx -23.94.
\]