Question

question 11 of 25
use the quadratic formula to find the solutions to the quadratic equation
below.
$x^2 - 5x - 4 = 0$

a. $x = \frac{5 \pm \sqrt{41}}{2}$

b. $x = \frac{-5 \pm \sqrt{41}}{2}$

c. $x = \frac{5 \pm \sqrt{29}}{2}$

d. $x = \frac{-5 \pm \sqrt{29}}{2}$

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 \(x^2-5x - 4=0\), we have \(a = 1\), \(b=-5\), \(c=-4\).

Step2: Identify coefficients

Here, \(a = 1\), \(b=-5\), \(c = - 4\).

Step3: Substitute into formula

Substitute \(a\), \(b\), \(c\) into the quadratic formula:
\(x=\frac{-(-5)\pm\sqrt{(-5)^2-4\times1\times(-4)}}{2\times1}\)
Simplify the numerator:
First, \(-(-5)=5\).
Then, \((-5)^2-4\times1\times(-4)=25 + 16=41\).
So, \(x=\frac{5\pm\sqrt{41}}{2}\)

Answer:

A. \(x=\frac{5\pm\sqrt{41}}{2}\)