# How do you find the solution to the quadratic equation Y=0.05x^2 + 1.1x?

Jun 21, 2018

Solving fo $x$ when $y = 0 \to x \text{ intercepts}$

$x = 0 , x = - 22$

#### Explanation:

Given: $y = 0.05 {x}^{2} + 1.1 x + 0$

$\textcolor{b l u e}{\text{The more efficient - it is a matter of spotting it.}}$

Set $y = 0 = 0.05 {x}^{2} + 1.1 x$

Factor out $x \to 0 = x \left(0.05 x + 1.1\right)$

Set $0.05 x + 1.1 = 0 \implies 0.05 x = - 1.1$

$x = - \frac{1.1}{0.05} = - \frac{110}{5} = - 22$

$x = 0 \mathmr{and} - 22$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The inefficient way: -Using the formula}}$

Using 0 as a place keeper

$y = a {x}^{x} + b x + c \to \text{ at y=0 we have } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case a=0.05; b=1.1; c=0

$x = \frac{- 1.1 \pm \sqrt{{\left(1.1\right)}^{2} - 4 \left(0.05\right) \left(0\right)}}{2 \left(0.05\right)}$

Anything times zero is zero so we have:

$x = \frac{- 1.1 \pm \sqrt{{\left(1.1\right)}^{2}}}{2 \left(0.05\right)}$

$x = - \frac{1.1}{0.1} \pm \frac{1.1}{0.1}$

$x = - 11 \pm 11$

$x = 0 , x = - 22$

Thus we have in factored form: $y = x \left(x + 22\right)$