Question

Let $f\left(x\right)=3a{x}^2-4bx+c(a,b,c\in R,a\ne 0)$ where $a,b,c$ are in $A.P$. Then the equation $f\left(x\right)=0$ has

• A. No real solution.
• B. Two unequal real roots.
• C. Sum of roots always negative.
• D. Product of roots always positive.

Answer 1

Answer: (B)

Two unequal real roots.

Since $a,b,c$ are in A.P., so,

$2b=a+c$

$4b^2={\left(a+c\right)}^2$

The discriminant of the given function$f\left(x\right)=3a{x}^2-4bx+c$ is,

$D=16b^2-12ac$

$=4{\left(a+c\right)}^2-12ac$

$=4\left[\left(a^2+c^2+2ac\right)-3ac\right]$

$=4\left(a^2+c^2-ac\right)$

$=4\left(a^2+c^2-2ac+ac\right)$

$=4\left({\left(a-c\right)}^2+ac\right)$

Case 1: If $a$ and$c$ are of opposite signs, then, $D=(+)ve$.

Case 2: If $a$ and$c$ are of same signs, then, $D=(+)ve$.

This shows that $f\left(x\right)=0$ has two unequal real roots.