Question

If roots of the equation $(a-b)x^2+(c-a)x+(b-c)=0,a\ne b\ne c$ are equal, then $a,b,c$ are in

• A. $A.P$
• B. $H.P$
• C. $G.P$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

We have,

Given equation is

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

On comparing that,

$A{x}^2+Bx+C=0$

Now,

$A=(a-b)$

$B=(c-a)$

$C=(b-c)$

Roots are equal

Then,

$D=0$

$B^2-4AC=0$

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

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

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

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

$\Rightarrow {(c+a)}^2=4b(a-b+c)$

Hence, this is the answer