Use inspection to solve the equation x^2=36/169, then select the correct solution set below. If there are no real solutions, select “There are no real solutions.”(1 point)

To solve the equation \( x^2 = \frac{36}{169} \), we take the square root of both sides.

This gives us:

\[ x = \pm \sqrt{\frac{36}{169}} \]

Calculating the square root of the fraction:

\[ \sqrt{\frac{36}{169}} = \frac{\sqrt{36}}{\sqrt{169}} = \frac{6}{13} \]

Thus, the solutions are:

\[ x = \frac{6}{13} \quad \text{and} \quad x = -\frac{6}{13} \]

So, the correct solution set is:

\[ \left{ -\frac{6}{13}, \frac{6}{13} \right} \]

The correct choice from the responses is:

\({−\frac{6}{13}, \frac{6}{13}}\)

Solve the following quadratic equation using square roots: (x−18)^2=196.(1 point)
Responses

​ {−4,−32}
​ left brace negative 4 comma negative 32 right brace

{32,4}
left brace 32 comma 4 right brace

{21,5}
left brace 21 comma 5 right brace

{214,−178}

To solve the equation \((x - 18)^2 = 196\) using square roots, we start by taking the square root of both sides:

\[ x - 18 = \pm \sqrt{196} \]

Calculating the square root of 196 gives us:

\[ \sqrt{196} = 14 \]

This leads to two equations:

1. \(x - 18 = 14\)
2. \(x - 18 = -14\)

Now, we solve each equation:

1. For \(x - 18 = 14\): \[ x = 14 + 18 = 32 \]

2. For \(x - 18 = -14\): \[ x = -14 + 18 = 4 \]

Thus, the solutions are \(x = 32\) and \(x = 4\).

The correct solution set is:

\({32, 4}\)

Solve the following quadratic equation using square roots: (x+6)^2/2=8.(1 point)
Responses

{−1.17,−6.83}
left brace negative 1.17 comma negative 6.83 right brace

​ {−0.34,−11.66}
​ left brace negative 0.34 comma negative 11.66 right brace

{−2,−10}
left brace negative 2 comma negative 10 right brace

{10,−22}

To solve the equation \(\frac{(x + 6)^2}{2} = 8\), we start by eliminating the fraction. We can do this by multiplying both sides by 2:

\[ (x + 6)^2 = 16 \]

Next, we take the square root of both sides:

\[ x + 6 = \pm \sqrt{16} \]

Calculating the square root gives:

\[ \sqrt{16} = 4 \]

This leads to two equations:

1. \(x + 6 = 4\)
2. \(x + 6 = -4\)

Now, we solve each equation:

1. For \(x + 6 = 4\): \[ x = 4 - 6 = -2 \]

2. For \(x + 6 = -4\): \[ x = -4 - 6 = -10 \]

Thus, the solutions are \(x = -2\) and \(x = -10\).

The correct solution set is:

\({-2, -10}\)

Solve the following quadratic equation using square roots: (x+36)^2−38=251.(1 point)
Responses

{−19,−53}
left brace negative 19 comma negative 53 right brace

{53,19}
left brace 53 comma 19 right brace

{253,−325}
left brace 253 comma negative 325 right brace

{17.84,−13.84}

To solve the equation \((x + 36)^2 - 38 = 251\), we first isolate the squared term.

1. Add 38 to both sides: \[ (x + 36)^2 = 251 + 38 \] \[ (x + 36)^2 = 289 \]

2. Next, take the square root of both sides: \[ x + 36 = \pm \sqrt{289} \]

Calculating the square root gives: \[ \sqrt{289} = 17 \]

So we have two equations to solve:

1. \(x + 36 = 17\)
2. \(x + 36 = -17\)

Now solve each equation:

1. For \(x + 36 = 17\): \[ x = 17 - 36 = -19 \]

2. For \(x + 36 = -17\): \[ x = -17 - 36 = -53 \]

Thus, the solutions are \(x = -19\) and \(x = -53\).

The correct solution set is:

\({-19, -53}\)