Question

If roots of quadratic equation $(b-c)x^2+(c-a)x+(a-b)=0$ are real and equal then prove that $2b=a+c$.

Answer 1

If roots of a quadratic equation are equal, then discriminant of the quadratic equation is 0

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

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

Comparing with

$a{x}^2+bx+c=0$

Here, $a=(b-c),b=(c-a)andc=(a-b)$

So,

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

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

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

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

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

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

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

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

$\Rightarrow 2b=c+a$