Question

Show how many solutions are in the solution set for 2(x−5)exponent2 +2=20 .(1 point)
Responses

one
one

two
two

zero
zero

infinitely many

Answer 1

To solve the equation \( 2(x-5)^2 + 2 = 20 \), we can start by isolating the quadratic term.

1. Subtract 2 from both sides:

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

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

2. Divide both sides by 2:

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

\[
(x-5)^2 = 9
\]

3. Take the square root of both sides:

\[
x - 5 = 3 \quad \text{or} \quad x - 5 = -3
\]

4. Solve for \( x \):

\[
x = 8 \quad \text{or} \quad x = 2
\]

Thus, there are **two solutions**: \( x = 8 \) and \( x = 2 \).

Therefore, the answer is:

**two**.