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

Question

2660 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-28T15:27:16

See below.

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

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

$x = -(bpm(a+b))/(2a)= {(1/2),(-(1/2+b/a)):}$

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

For a certain equation, a, b, ca,b,ca, b, c belongs to real numbers. ab > 0ab>0ab > 0 and a+2b+4c=0a+2b+4c=0a+2b+4c=0. Then, the equation ax^2+bx+c=0ax2+bx+c=0ax^2+bx+c=0 has what kind of roots (one +ve & one -ve, both +ve or both -ve) ? Thank you!