# For a certain equation, a, b, c belongs to real numbers. ab > 0 and a+2b+4c=0. Then, the equation ax^2+bx+c=0 has what kind of roots (one +ve & one -ve, both +ve or both -ve) ? Thank you!

Nov 28, 2016

See below.

#### Explanation:

Putting $c = - \frac{a + 2 b}{4}$ into

$a {x}^{2} + b x + c = 0$ and solving for $x$ we obtain

$x = - \frac{b \pm \left(a + b\right)}{2 a} = \left\{\begin{matrix}\frac{1}{2} \\ - \left(\frac{1}{2} + \frac{b}{a}\right)\end{matrix}\right.$

so if $a b > 0$ we have two real distinct roots, with opposite signs.