Show how many solutions are in the solution set for the equation 3x2=21.(1 point) Responses zero zero two two one one infinitely many

Answer 1

To solve the equation \( 3x^2 = 21 \), we first divide both sides by 3:

\[ x^2 = \frac{21}{3} = 7 \]

Next, we take the square root of both sides:

\[ x = \pm \sqrt{7} \]

This means there are two solutions:

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

Therefore, the number of solutions in the solution set for the equation \( 3x^2 = 21 \) is two.

Answer 2

Show how many solutions are in the solution set for 2(x−5)2+2=20.(1 point) Responses one one infinitely many infinitely many two two zero

Answer 3

To solve the equation \( 2(x-5)^2 + 2 = 20 \), we first isolate the term with the squared expression.

1. Subtract 2 from both sides:

   \[ 2(x-5)^2 = 20 - 2 \] \[ 2(x-5)^2 = 18 \]

2. Divide both sides by 2:

   \[ (x-5)^2 = \frac{18}{2} = 9 \]

3. Next, take the square root of both sides:

   \[ x - 5 = \pm 3 \]

This leads to two equations:

• \( x - 5 = 3 \)
• \( x - 5 = -3 \)

Now, we solve these equations:

1. For \( x - 5 = 3 \): \[ x = 3 + 5 = 8 \]

2. For \( x - 5 = -3 \): \[ x = -3 + 5 = 2 \]

Thus, the solutions are \( x = 8 \) and \( x = 2 \).

Therefore, the number of solutions in the solution set for the equation \( 2(x-5)^2 + 2 = 20 \) is two.

Answer 4

Solve x2=16/25 by inspection. There are two real solutions. Enter the lesser number first. Leave the answers in simplest fraction form

Answer 5

To solve the equation \( x^2 = \frac{16}{25} \) by inspection, we recognize that we can take the square root of both sides.

Taking the square root, we get:

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

Calculating the square root of the fraction:

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

Thus, the two real solutions are \( \frac{4}{5} \) and \( -\frac{4}{5} \).

Arranging these solutions in order, with the lesser number first, we have:

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

So the final answers are:

\(-\frac{4}{5}, \frac{4}{5}\)