Question

question 5 of 25
use the quadratic formula to find the solutions to the quadratic equation below.
$2x^2 - 5x + 5 = 0$

a. $x = \frac{5\pm i\sqrt{15}}{4}$

b. $x = \frac{5\pm \sqrt{15}}{4}$

c. $x = \frac{5\pm i\sqrt{53}}{4}$

d. $x = \frac{5\pm \sqrt{53}}{4}$

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 \(2x^2-5x + 5=0\), we have \(a = 2\), \(b=- 5\), \(c = 5\).

Step2: Calculate discriminant (\(b^2-4ac\))

Substitute the values of \(a\), \(b\), and \(c\) into the discriminant formula:
\(b^2-4ac=(-5)^2-4\times2\times5=25 - 40=- 15\)

Step3: Substitute into quadratic formula

Since \(\sqrt{-15}=\sqrt{15}\times\sqrt{-1}=i\sqrt{15}\) (where \(i\) is the imaginary unit, \(i^2=-1\)), substitute into the quadratic formula:
\(x=\frac{-(-5)\pm\sqrt{-15}}{2\times2}=\frac{5\pm i\sqrt{15}}{4}\)

Answer:

A. \(x=\frac{5\pm i\sqrt{15}}{4}\)