If the quadratic equation a(b-c) x2+ b(c-a) x+c(a-b)=0, where a, b, c are distinct real numbers and a b c ≠ 0, has equal roots, then a, b, c are in

Question

If the quadratic equation a(b-c) x2+ b(c-a) x+c(a-b)=0, where a, b, c are distinct real numbers and a b c ≠ 0, has equal roots, then a, b, c are in (A) A.P. (B) G.P. (C) H.P. (D) A.G.P.

Tardigrade Admin · Accepted Answer

As 1 is a root of the equation and equation has equal roots, the other root must be 1 . Thus we have

1 = product of roots = \frac{c ( a - b )}{a ( b - c )}

\Rightarrow a (b - c) = c (a - b)

\Rightarrow b = \frac{2 a c}{a + c}

\Rightarrow a, b, c are in H.P.

Q. If the quadratic equation a(b−c)x2+ b(c−a)x+c(a−b)=0, where a,b,c are distinct real numbers and abc=0, has equal roots, then a,b,c are in

A.P.

G.P.

H.P.

A.G.P.

As 1 is a root of the equation and equation has equal roots, the other root must be 1 . Thus we have 1= product of roots =a(b−c)c(a−b)​ ⇒a(b−c)=c(a−b) ⇒b=a+c2ac​ ⇒a,b,c are in H.P.

Q. If the quadratic equation $a (b - c) x^{2} +$ $b (c - a) x + c (a - b) = 0$ , where $a, b, c$ are distinct real numbers and $ab c \neq = 0$ , has equal roots, then $a, b, c$ are in

As 1 is a root of the equation and equation has equal roots, the other root must be 1 . Thus we have
$1 = product of roots = \frac{c ( a - b )}{a ( b - c )}$
$\Rightarrow a (b - c) = c (a - b)$
$\Rightarrow b = \frac{2 a c}{a + c}$
$\Rightarrow a, b, c are in H.P.$