(x-6)^2=51

Answer 1

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 2

(x+16)^2/9=7

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:

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