Question

If the roots of $a{x}^2+b=0$ are real, where $a\ne 0,b\ne 0$ then :

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

Answer 1

Answer: (A), (C)

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

Since the roots of the given equation, $a{x}^2+b=0$ are real

$\therefore \text{discriminant}=B^2-4AC\ge 0$

Here, $B=0,A=a,C=b$ by comparing the general quadratic equation $A{x}^2+Bx+C=0$

$\Rightarrow -4ab>0$ ($4ab\ne 0$ because neither $a\ne 0$ nor $b\ne 0$)

$\Rightarrow ab<0$

Therefore, for the above inequality to hold Either $a<0$ and $b>0$ or $a>0$ and $b<0$.