Question

If $b^2\ge 4ac$ for the equation $a{x}^4+b{x}^2+c=0,$then all the roots of the equation will be real, if,

• A. $b>0,a<0,c>0$
• B. $b<0,a>0,c>0$
• C. $b>0,a>0,c>0$
• D. $b>0,a>0,c<0$

Answer 1

The correct option is B

$b<0,a>0,c>0$

Finding whether the roots of the equation will be real positive if:

Put ${\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathit{y}$

$\Rightarrow a{y}^2+by+c=0\to \left(1\right)$

The given equation will have four real roots if $\left(1\right)$ has have two non-negative roots.This happen if

$\Rightarrow$ $-\frac{b}{a}\ge 0$, $af\ge 0$,$b^2-4ac\ge 0$

$\Rightarrow$$-ab\ge 0,$$ac\ge 0$ ($b^2-4ac\ge 0$ is given)

Then $a$ and $b$must have opposite sign.whereas, $a,-b$ and $c$ must have the same sign.

$\Rightarrow$$a>0,b<0,c>0$ or $a<0,b>0,c<0$.

Hence option(B) is correct.