Question

Which equation has two real solutions?

A. x2 = –100
B. 5x2 = 1
C. 6x2 + 17 = 11
D. 7(x2 + 6) = 42

Answer 1

To determine which equation has two real solutions, we can analyze each option one by one.

**A.** \( x^2 = -100 \)

This equation has no real solutions because the square of a real number cannot be negative.

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**B.** \( 5x^2 = 1 \)

To solve for \( x \):

\[
x^2 = \frac{1}{5}
\]

Taking the square root of both sides gives:

\[
x = \pm \sqrt{\frac{1}{5}} = \pm \frac{1}{\sqrt{5}} \approx \pm 0.447
\]

This equation has two real solutions.

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**C.** \( 6x^2 + 17 = 11 \)

First, simplify the equation:

\[
6x^2 = 11 - 17
\]
\[
6x^2 = -6 \quad \Rightarrow \quad x^2 = -1
\]

This equation has no real solutions, as again, the square of a real number cannot be negative.

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**D.** \( 7(x^2 + 6) = 42 \)

First, simplify the equation:

\[
x^2 + 6 = \frac{42}{7} = 6 \quad \Rightarrow \quad x^2 = 6 - 6 = 0
\]

This equation has exactly one real solution \( x = 0 \).

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In summary, the only equation among the options that has two real solutions is:

**B.** \( 5x^2 = 1 \)