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
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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
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◘ 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.
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 –
[tex] \\ \implies \large { \boxed{ \bold{D = 0}}} \\ [/tex]
• And –
[tex] \\ \implies \large { \boxed{ \bold{D = {b}^{2} - 4ac }}} \\ [/tex]
• So that –
[tex] \\ \implies \large { \boxed{ \bold{ {b}^{2} - 4ac = 0}}} \\ [/tex]
• Here –
[tex] \\ \: \: { \huge{.}} \: \: \: \: { \bold{a = (a - b)}} \\ [/tex]
[tex] \\ \: \: { \huge{.}} \: \: \: \: { \bold{b = (c - a)}} \\ [/tex]
[tex] \\ \: \: { \huge{.}} \: \: \: \: { \bold{c = (b - c)}} \\ [/tex]
• So that –
[tex] \\ \implies { \bold{ {(c - a)}^{2} - 4(a - b)(b - c) = 0}} \\ [/tex]
[tex] \\ \implies { \bold{{ {c}^{2} + {a}^{2} -2 ac= 4(ab - ac - {b}^{2} +bc )}}} \\ [/tex]
[tex] \\ \implies { \bold{{ {c}^{2} + {a}^{2} - 2ac - 4ab + 4ac + 4 {b}^{2} - 4 bc= 0}}} \\ [/tex]
[tex] \\ \implies { \bold{{ {c}^{2} + {a}^{2} - 4ab + 2ac + 4 {b}^{2} - 4 bc= 0}}} \\ [/tex]
• We should write this as –
[tex] \\ \implies { \bold{ {(a - 2b + c)}^{2} = 0}} \\ [/tex]
[tex] \\ \implies { \bold{ {(a - 2b + c)} = 0}} \\ [/tex]
[tex] \\ \implies { \bold{ a + c = 2b}} \\ [/tex]
[tex] \\ \implies \large { \boxed{ \bold{b = \dfrac{a + c}{2} }}} \\ [/tex]
• This condition possible if a , b and c are in A.P.
• Hence, Option (A) is correct .