Question

If $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has equal root, then $a,b,c$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is A A.P.

Roots of quadratic equation when it is equal

$b^2-4ac=0$

Given in the equation

$(b-c)x^2+(c-a)x+(a-b)=0$

Here if we put the values insides the quadratic equation $b^2-4ac:-$

$b=(c-a)$

$a=(b-c)$

$c=(a-b)$

Discriminant $D=b^2-4ac=0$

$\Rightarrow (c-a{)}^{2}4(a-b\left)\right(b-c)=0$

$\Rightarrow (c^2+a^2-2ac)-4(ab-b^2-ac+bc)=0$

$\Rightarrow c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0$

$\Rightarrow a^2+4b^2+c^2-4ab-4bc+2ac=0$

$\Rightarrow (a-2b+c{)}^2=0$

$\Rightarrow a-2b+c=0$

$\Rightarrow a+c=2b$

$\Rightarrow (b-a)=(c-b)$ [We can write the expression

$a+c=2baas(b-a)=(c-b)$

So the required condition for $a,b,c$ therefore

$a,b,c$ are in A.P [Proved]