Question

If $b^2\ge 4ac$ for the equation $a{x}^4+b{x}^2+c=0$, then all 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

Answer: (B), (D)

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

$x^2=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
It is already given that $b^2>4ac$.
$x^2=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x^2=\frac{-b-\sqrt{b^2-4ac}}{2a}$
For both roots to be real,
Either $b>0$ and $a,c<0$ or $b<0$ and $a,c>0$.