Question

1. what is the solution to the equation $2x^{2}+4x - 5 = 0$ in simplified radical form?

a. $x = \frac{-4\pm\sqrt{56}}{4}$
b. $x = \frac{4\pm\sqrt{56}}{4}$
c. $x = \frac{-4\pm2\sqrt{14}}{4}$
d. $x = \frac{-2\pm\sqrt{14}}{2}$

Explanation:

Step1: Recall Quadratic Formula

For a quadratic equation \(ax^2 + bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = 4\), \(c=-5\).

Step2: Calculate Discriminant

Discriminant \(D=b^2 - 4ac=(4)^2-4\times2\times(-5)=16 + 40 = 56\).

Step3: Substitute into Quadratic Formula

\(x=\frac{-4\pm\sqrt{56}}{2\times2}=\frac{-4\pm2\sqrt{14}}{4}\) (since \(\sqrt{56}=\sqrt{4\times14}=2\sqrt{14}\)).

Step4: Simplify the Expression

Divide numerator and denominator by 2: \(x=\frac{-2\pm\sqrt{14}}{2}\).

Answer:

D. \(x = \frac{-2\pm\sqrt{14}}{2}\)