# How do you use the quadratic formula to solve 4x^2+16x+15=0?

Nov 9, 2016

See explanation...

#### Explanation:

$4 {x}^{2} + 16 x + 15 = 0$

is in the form:

$a {x}^{2} + b x + c = 0$

with $a = 4$, $b = 16$ and $c = 15$

The roots are given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \left(4\right) \left(15\right)}}{2 \cdot 4}$

$\textcolor{w h i t e}{x} = \frac{- 16 \pm \sqrt{256 - 240}}{8}$

$\textcolor{w h i t e}{x} = \frac{- 16 \pm \sqrt{16}}{8}$

$\textcolor{w h i t e}{x} = \frac{- 16 \pm 4}{8}$

$\textcolor{w h i t e}{x} = \frac{- 4 \pm 1}{2}$

That is:

$x = - \frac{5}{2}$ or $x = - \frac{3}{2}$

$\textcolor{w h i t e}{}$
Footnote

In this particular example, I would probably have favoured an AC method rather than the quadratic formula.

Given:

$4 {x}^{2} + 16 x + 15 = 0$

Find a pair of factors of $A C = 4 \cdot 15 = 60$ with sum $B = 16$

The pair $10 , 6$ works.

Use this pair to split the middle term and factor by grouping:

$0 = 4 {x}^{2} + 16 x + 15$

$\textcolor{w h i t e}{0} = \left(4 {x}^{2} + 10 x\right) + \left(6 x + 15\right)$

$\textcolor{w h i t e}{0} = 2 x \left(2 x + 5\right) + 3 \left(2 x + 5\right)$

$\textcolor{w h i t e}{0} = \left(2 x + 3\right) \left(2 x + 5\right)$

Hence roots $x = - \frac{3}{2}$ and $x = - \frac{5}{2}$

Note that the AC method is only applicable if the quadratic has rational zeros, whereas the quadratic formula will always give you the solution, even if the roots are Complex.