Solve x^2=36

To solve the equation \( x^2 = 36 \) by inspection, we can find the values of \( x \) that satisfy this equation.

Taking the square root of both sides gives us:

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

Since \( \sqrt{36} = 6 \), the solutions are:

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

The lesser number is \( -6 \). Therefore, the two real solutions are:

\[ -6, 6 \]

Solve x^2=16/25 by inspection. There are two real solutions. Enter the lesser number first. Leave the answers in simplest fraction form.

To solve the equation \( x^2 = \frac{16}{25} \) by inspection, we can find the values of \( x \) that satisfy this equation.

Taking the square root of both sides gives us:

\[ x = \pm \sqrt{\frac{16}{25}} \]

Calculating the square root, we get:

\[ x = \pm \frac{\sqrt{16}}{\sqrt{25}} = \pm \frac{4}{5} \]

The two real solutions are:

\[ -\frac{4}{5} \quad \text{and} \quad \frac{4}{5} \]

Arranging them with the lesser number first, the solutions are:

\[ -\frac{4}{5}, \frac{4}{5} \]

Solve the following quadratic equation using square roots: (x+3)^2=36.

To solve the quadratic equation \( (x + 3)^2 = 36 \) using square roots, we can follow these steps:

1. Take the square root of both sides of the equation: \[ x + 3 = \pm \sqrt{36} \]

2. Since \( \sqrt{36} = 6 \), we have: \[ x + 3 = 6 \quad \text{or} \quad x + 3 = -6 \]

3. Now, solve for \( x \) in both cases.

   • For the first case: \[ x + 3 = 6 \] Subtract 3 from both sides: \[ x = 6 - 3 = 3 \]

   • For the second case: \[ x + 3 = -6 \] Subtract 3 from both sides: \[ x = -6 - 3 = -9 \]

Thus, the solutions to the equation \( (x + 3)^2 = 36 \) are: \[ x = 3 \quad \text{and} \quad x = -9 \]

Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x−6)^2=51.

To solve the quadratic equation \( (x - 6)^2 = 51 \) using square roots, we can follow these steps:

1. Take the square root of both sides: \[ x - 6 = \pm \sqrt{51} \]

2. Now calculate \( \sqrt{51} \). It's approximately: \[ \sqrt{51} \approx 7.1414 \]

3. Now we can set up the two equations: \[ x - 6 = 7.1414 \quad \text{or} \quad x - 6 = -7.1414 \]

4. Solve for \( x \) in both cases:

   • For the first case: \[ x - 6 = 7.1414 \] Add 6 to both sides: \[ x \approx 7.1414 + 6 \approx 13.1414 \]

   • For the second case: \[ x - 6 = -7.1414 \] Add 6 to both sides: \[ x \approx -7.1414 + 6 \approx -1.1414 \]

5. Rounding to the nearest hundredth:

   • For the first case: \( x \approx 13.14 \)
   • For the second case: \( x \approx -1.14 \)

Thus, the solutions to the equation \( (x - 6)^2 = 51 \) rounded to the nearest hundredth are: \[ x \approx 13.14 \quad \text{and} \quad x \approx -1.14 \]

Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x+16)^2/9=7
.

To solve the quadratic equation \(\frac{(x+16)^2}{9} = 7\), we can follow these steps:

1. Multiply both sides by 9 to eliminate the fraction: \[ (x + 16)^2 = 63 \]

2. Take the square root of both sides: \[ x + 16 = \pm \sqrt{63} \]

3. Calculate \(\sqrt{63}\): \[ \sqrt{63} \approx 7.937 \]

4. Set up the two equations: \[ x + 16 = 7.937 \quad \text{or} \quad x + 16 = -7.937 \]

5. Solve for \(x\) in both cases:

   • For the first case: \[ x + 16 = 7.937 \] Subtract 16 from both sides: \[ x = 7.937 - 16 \approx -8.063 \]

   • For the second case: \[ x + 16 = -7.937 \] Subtract 16 from both sides: \[ x = -7.937 - 16 \approx -23.937 \]

6. Round to the nearest hundredth:

   • For the first case, \( x \approx -8.06 \)
   • For the second case, \( x \approx -23.94 \)

Thus, the solutions to the equation \(\frac{(x + 16)^2}{9} = 7\) rounded to the nearest hundredth are: \[ x \approx -8.06 \quad \text{and} \quad x \approx -23.94 \]

Use inspection to solve the equation x^2=81, then select the complete solution set below. If there are no real solutions, select “There are no real solutions.”(1 point)
Responses

{−9}
left brace negative 9 right brace

{9}
left brace 9 right brace

There are no real solutions.
There are no real solutions.

{−9,9}

To solve the equation \( x^2 = 81 \) by inspection, we can determine the values of \( x \) that satisfy this equation.

Taking the square root of both sides gives us:

\[ x = \pm \sqrt{81} \]

Since \( \sqrt{81} = 9 \), the solutions are:

\[ x = 9 \quad \text{and} \quad x = -9 \]

Thus, the complete solution set is:

\[ {-9, 9} \]

So the correct response is:

{−9, 9}