Question

If the roots of the equation (b - c)$x^2$ + (c - a)x + (a - b) = 0 be equal, then prove that a, b, c are in arithmetic progression.

Answer 1

Given, (b-c)$x^2$ + (c - a)x + (a - b) = 0
Comparing given equation with general form $A{x}^3+Bx+C=0$

We get, $A=(b-c),B=(c-a),D=(a-b)$

If roots are equal then discriminant is equal to zero.
That is, $D=B^2-4AC=0$

Therefore,

$(c-a{)}^2=4(a-b\left)\right(b-c)$
$c^2+a^2-2ac$ = 4 {ab - ac $-b^2$ + bc}
$c^2+a^2-2ac$ = 4ab - 4ac - $4b^2$ + 4bc
$a^2+4b^2+c^2+2ac-4ab-4bc=0$
$(a-2b-c{)}^2$ = 0
$a-2b+c=0$
$a+c=2b$
$b=\frac{a+c}{2}$
This means $a,b$ & $c$ are in A.P.

Hence, proved.