Question

If $a,b,c$ are in $HP$, then the expression $a(b-c)x^2+b(c-a)x+c(a-b)$

• A. has real and distinct factors
• B. is a perfect square
• C. has no real factor
• D. none of these

Answer 1

The correct option is B is a perfect square

Since, $a,b,c$ are in H.P.
Therefore, $b=\frac{2ac}{a+c}$. ....(1)
Consider $a(b-c)x^2+b(c-a)x+c(a-b)=0$.
Now, $D=b^2{(c-a)}^2-4ac(b-c)(a-b)=b^2[{(c+a)}^2-4ac]+4ac[b^2-b(a+c)+ac]$
$\Rightarrow D=\frac{4a^2{c}^2[{(c+a)}^2-4ac]}{{(c+a)}^2}+4ac[\frac{4a^2{c}^2}{{(a+c)}^2}-\frac{2ac(a+c)}{a+c}+ac]$ ...[ from (1) ]
$\Rightarrow D=0$
Since, the roots of the quadratic equation are equal.
Therefore, $a(b-c)x^2+b(c-a)x+c(a-b)$ is a perfect square.
Ans: B