Question

If roots of the equation (a – b)x^2 + (c – a)x + (b – c) = 0 are equal, then a, b, c are in
A) A.P.
B) H.P.
C) G.P.
D) None of these​

Answer 1

◘ Given ◘

Roots of the equation (a - b)x² + (c - a)x + (b - c) = 0 are equal.

◘ To Find ◘

If a, b, c are in

☛ A) A.P.

☛ B) H.P.

☛ C) G.P.

☛ D) None of these​

◘ Solution ◘

(a – b)x² + (c – a)x + (b – c) = 0

• As roots are equal so,

b²  - 4ac = 0

⇒ (c - a)² - 4(a - b)(b - c) = 0

⇒ (c - a)² - 4ab + 4b² + 4ac - 4bc = 0

⇒ (c - a)² + 4ac - 4b(c + a) + 4b² = 0

⇒ (c + a)² - 2.(2b)(c + a) + (2b)² = 0

⇒ [c + a - 2b]² = 0

⇒ c + a - 2b = 0

⇒ c + a = 2b

Hence, a, b & c are in A.P.

____________________

ALTERNATE :-

∵ Sum of coefficients = 0,

Hence one root is 1 and the other root is (b - c) / (a - b).

A/q,

Both the zeros are equal. So,

(b - c) / (a - b) = 1

⇒ b - c = a - b

⇒ 2b = a + c

Hence a, b, c are in A.P.

Therefore, the answer is (A) A.P.

Answer 2

GIVEN :–

• Roots of the equation (a – b)x² + (c – a)x + (b – c) = 0 are equal.

TO FIND :–

• Relation between a , b , c = ?

SOLUTION :–

• If a quadratic equation ax² + bx + c = 0 have equal roots then Discriminant –

$\\ \implies \large { \boxed{ \bold{D = 0}}}$

• And –

$\\ \implies \large { \boxed{ \bold{D = {b}^{2} - 4ac }}}$

• So that –

$\\ \implies \large { \boxed{ \bold{ {b}^{2} - 4ac = 0}}}$

• Here –

$\\ \: \: { \huge{.}} \: \: \: \: { \bold{a = (a - b)}}$

$\\ \: \: { \huge{.}} \: \: \: \: { \bold{b = (c - a)}}$

$\\ \: \: { \huge{.}} \: \: \: \: { \bold{c = (b - c)}}$

• So that –

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

$\\ \implies { \bold{{ {c}^{2} + {a}^{2} -2 ac= 4(ab - ac - {b}^{2} +bc )}}}$

$\\ \implies { \bold{{ {c}^{2} + {a}^{2} - 2ac - 4ab + 4ac + 4 {b}^{2} - 4 bc= 0}}}$

$\\ \implies { \bold{{ {c}^{2} + {a}^{2} - 4ab + 2ac + 4 {b}^{2} - 4 bc= 0}}}$

• We should write this as –

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

$\\ \implies { \bold{ {(a - 2b + c)} = 0}}$

$\\ \implies { \bold{ a + c = 2b}}$

$\\ \implies \large { \boxed{ \bold{b = \dfrac{a + c}{2} }}}$

• This condition possible if a , b and c are in A.P.

• Hence, Option (A) is correct .