Question

Which of the following is a solution to x + x ^ 2 - 1?

(A)-1

(B) 0

(C) ½

(D) 1

Answer 1

To find which of the options is a solution to the equation \( x + x^2 - 1 = 0 \), we can substitute each option into the equation and check if it satisfies the equation.

1. **Option (A): \( x = -1 \)**
\[
-1 + (-1)^2 - 1 = -1 + 1 - 1 = -1 \quad (\text{not } 0)
\]

2. **Option (B): \( x = 0 \)**
\[
0 + 0^2 - 1 = 0 + 0 - 1 = -1 \quad (\text{not } 0)
\]

3. **Option (C): \( x = \frac{1}{2} \)**
\[
\frac{1}{2} + \left(\frac{1}{2}\right)^2 - 1 = \frac{1}{2} + \frac{1}{4} - 1
\]
Converting to a common denominator (4):
\[
\frac{2}{4} + \frac{1}{4} - \frac{4}{4} = \frac{2 + 1 - 4}{4} = \frac{-1}{4} \quad (\text{not } 0)
\]

4. **Option (D): \( x = 1 \)**
\[
1 + 1^2 - 1 = 1 + 1 - 1 = 1 \quad (\text{not } 0)
\]

None of the options seem to satisfy the equation \( x + x^2 - 1 = 0 \).

To double-check, let's actually solve the equation \( x + x^2 - 1 = 0 \):
Rearranging gives us:
\[
x^2 + x - 1 = 0
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 1, c = -1 \):
\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}
\]

Since the roots are:
\[
x = \frac{-1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{5}}{2}
\]

Neither of these roots matches any of the options provided. Therefore, none of the options (A, B, C, D) is a solution to the equation \( x + x^2 - 1 = 0 \).