Question

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

Answer 1

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

Answer 2

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 gives us two solutions:

1. \(x = \sqrt{7}\)
2. \(x = -\sqrt{7}\)

Therefore, there are **two solutions** in the solution set for the equation \(3x^2 = 21\).

Answer 3

Let's solve the equation \(2(x - 5)^2 + 2 = 20\).

First, we can subtract 2 from both sides:

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

Next, divide both sides by 2:

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

Now we take the square root of both sides:

\[
x - 5 = \pm 3
\]

This gives us two cases to solve for \(x\):

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

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

Thus, the solutions are \(x = 8\) and \(x = 2\). Therefore, there are **two solutions** in the solution set for the equation \(2(x - 5)^2 + 2 = 20\).