Question

question 8 of 10
use the quadratic formula to find both solutions to the quadratic equation
given below.
$2x^2 + x - 1 = 0$

a. $x = \frac{-5}{2}$
b. $x = \frac{3 - \sqrt{8}}{4}$
c. $x = \frac{1}{2}$
d. $x = \frac{3 + \sqrt{8}}{4}$
e. $x = 1$
f. $x = -1$

Explanation:

Step1: Recall quadratic formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For the equation \(2x^2+x - 1=0\), we have \(a = 2\), \(b = 1\), \(c=- 1\).

Step2: Calculate discriminant

First, calculate the discriminant \(D=b^2-4ac=(1)^2-4\times2\times(-1)=1 + 8=9\).

Step3: Substitute into quadratic formula

Substitute \(a = 2\), \(b = 1\), \(D = 9\) into the quadratic formula:
\(x=\frac{-1\pm\sqrt{9}}{2\times2}=\frac{-1\pm3}{4}\)

Step4: Find two solutions

For the plus sign: \(x=\frac{-1 + 3}{4}=\frac{2}{4}=\frac{1}{2}\)
For the minus sign: \(x=\frac{-1-3}{4}=\frac{-4}{4}=-1\)

Answer:

C. \(x=\frac{1}{2}\), F. \(x = - 1\)