Question

question 6 of 10
use the quadratic formula to find both solutions to the quadratic equation
given below.
$4x^2 + 5x + 1 = 0$

a. $x = \frac{-3+\sqrt{7}}{8}$
b. $x = \frac{-3-\sqrt{7}}{8}$
c. $x = \frac{-3+\sqrt{-7}}{8}$
d. $x = -\frac{1}{4}$
e. $x = \frac{-3-\sqrt{-7}}{8}$
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 \(4x^2+5x + 1 = 0\), we have \(a = 4\), \(b = 5\), and \(c = 1\).

Step2: Calculate the Discriminant

First, calculate the discriminant \(D=b^2-4ac\). Substitute \(a = 4\), \(b = 5\), \(c = 1\) into the formula:
\(D=(5)^2-4\times4\times1=25 - 16=9\)

Step3: Apply Quadratic Formula

Now, substitute \(a = 4\), \(b = 5\), and \(D = 9\) (so \(\sqrt{D}=\sqrt{9} = 3\)) into the quadratic formula:
\(x=\frac{-5\pm3}{2\times4}=\frac{-5\pm3}{8}\)

Step4: Find the Two Solutions

• For the plus sign: \(x=\frac{-5 + 3}{8}=\frac{-2}{8}=-\frac{1}{4}\)
• For the minus sign: \(x=\frac{-5-3}{8}=\frac{-8}{8}=-1\)

Answer:

D. \(x = -\frac{1}{4}\), F. \(x = -1\)