# Question #efd42

Oct 26, 2016

Given quadratic equation of m is

$2 {m}^{2} - 16 m + 8 = 0$

$\implies {m}^{2} - 8 m + 4 = 0$

It being a quasratic equation it will have only two values that satisfy the equation. Let these are a and b.

So $\left(m - a\right) \left(m - b\right) = 0$ should be the qudratic equation.

Comparing these two equations
we have

$\left(m - a\right) \left(m - b\right) = {m}^{2} - 8 m + 4$

${m}^{2} - \left(a + b\right) m + a b = {m}^{2} - 8 m + 4$

Comparing LHS and RHS we get

$a + b = 8$

~~~~~~~~~~~~~~~~~~~~~~~~~~±~
$a {x}^{2} + b x + c = 0$,
if $\alpha \mathmr{and} \beta$ are two values of x that satisfy the equation then
$\alpha + \beta = - \frac{b}{a} \mathmr{and} \alpha \beta = \frac{c}{a}$