Using the quadratic formula, solve

[tex]\[ 4x^2 + 16x + 15 = 0 \][/tex]

to find the two possible values of \( x \). Give any decimal answers to 1 d.p.

The quadratic formula is

[tex]\[ ax^2 + bx + c = 0 \][/tex]

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]



Answer :

To solve the quadratic equation \(4x^2 + 16x + 15 = 0\) using the quadratic formula, follow these steps:

The general quadratic equation is given by:
[tex]\[ ax^2 + bx + c = 0 \][/tex]

Here, the coefficients are:
[tex]\[ a = 4, \quad b = 16, \quad c = 15 \][/tex]

The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

1. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{Discriminant} = 16^2 - 4 \cdot 4 \cdot 15 \][/tex]
[tex]\[ \text{Discriminant} = 256 - 240 \][/tex]
[tex]\[ \text{Discriminant} = 16 \][/tex]

2. Calculate the two possible values of \(x\) using the quadratic formula. We have:
[tex]\[ x = \frac{-b \pm \sqrt{\text{Discriminant}}}{2a} \][/tex]

First root (\( x_1 \)):
[tex]\[ x_1 = \frac{-b + \sqrt{\text{Discriminant}}}{2a} \][/tex]
[tex]\[ x_1 = \frac{-16 + \sqrt{16}}{2 \cdot 4} \][/tex]
[tex]\[ x_1 = \frac{-16 + 4}{8} \][/tex]
[tex]\[ x_1 = \frac{-12}{8} \][/tex]
[tex]\[ x_1 = -1.5 \][/tex]

Second root (\( x_2 \)):
[tex]\[ x_2 = \frac{-b - \sqrt{\text{Discriminant}}}{2a} \][/tex]
[tex]\[ x_2 = \frac{-16 - \sqrt{16}}{2 \cdot 4} \][/tex]
[tex]\[ x_2 = \frac{-16 - 4}{8} \][/tex]
[tex]\[ x_2 = \frac{-20}{8} \][/tex]
[tex]\[ x_2 = -2.5 \][/tex]

So, the two possible values of \(x\), rounded to 1 decimal place, are:
[tex]\[ x_1 = -1.5 \][/tex]
[tex]\[ x_2 = -2.5 \][/tex]