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Question: If the roots of \(a ( b - c ) x ^ { 2 } + b ( c - a ) x + c ( a - b ) = 0\) be equal, then *a*, *b...

If the roots of a(b−c)x2+b(c−a)x+c(a−b)=0a ( b - c ) x ^ { 2 } + b ( c - a ) x + c ( a - b ) = 0 be equal, then a, b, c are in

A

A.P.

B

G.P.

C

H.P.

D

None of these

Answer

H.P.

Explanation

Solution

As the roots are equal, discriminate = 0

⇒ {b(c−a)}2−4a(b−c)c(a−b)=0\{ b ( c - a ) \} ^ { 2 } - 4 a ( b - c ) c ( a - b ) = 0

⇒b2c2+a2b2−2ab2c−4a2bc+4a2c2+4ab2c−4abc2=0b ^ { 2 } c ^ { 2 } + a ^ { 2 } b ^ { 2 } - 2 a b ^ { 2 } c - 4 a ^ { 2 } b c + 4 a ^ { 2 } c ^ { 2 } + 4 a b ^ { 2 } c - 4 a b c ^ { 2 } = 0

⇒(b2c2+2ab2c+a2b2)=4ac{ab+bc−ac}\left( b ^ { 2 } c ^ { 2 } + 2 a b ^ { 2 } c + a ^ { 2 } b ^ { 2 } \right) = 4 a c \{ a b + b c - a c \}

⇒(ab+bc)2=4ac(ab+bc−ac)( a b + b c ) ^ { 2 } = 4 a c ( a b + b c - a c )

⇒ {b(a+c)}2=4abc(a+c)−4a2c2\{ b ( a + c ) \} ^ { 2 } = 4 a b c ( a + c ) - 4 a ^ { 2 } c ^ { 2 }

⇒ b2(a+c)2−2b(a+c)⋅2ac+(2ac)2=0b ^ { 2 } ( a + c ) ^ { 2 } - 2 b ( a + c ) \cdot 2 a c + ( 2 a c ) ^ { 2 } = 0

⇒ [b(a+c)−2ac]2=0[ b ( a + c ) - 2 a c ] ^ { 2 } = 0

∴b=2aca+cb = \frac { 2 a c } { a + c }

Thus, a, b, c are in H.P.