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Question: If \(\frac{x^{2} - bx}{ax–c} = \frac{\lambda - 1}{\lambda + 1}\)has roots equal in magnitude and opp...

If x2−bxax–c=λ−1λ+1\frac{x^{2} - bx}{ax–c} = \frac{\lambda - 1}{\lambda + 1}has roots equal in magnitude and opposite

in sign, then the value of l is –

A

a−ba+b\frac{a - b}{a + b}

B

a+ba−b\frac{a + b}{a - b}

C

c

D

1c\frac{1}{c}

Answer

a−ba+b\frac{a - b}{a + b}

Explanation

Solution

x2−bxax−c=λ−1λ+1\frac{x^{2} - bx}{ax - c} = \frac{\lambda - 1}{\lambda + 1}

Ž l x2 – (bl + al + b – a) x + cl – c = 0

sum of root = 0 Ž bλ+aλ+b−aλ\frac{b\lambda + a\lambda + b - a}{\lambda}= 0

Ž l = a−ba+b\frac{a - b}{a + b}