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Question: If the roots of the equation \(\frac{x^{2} - bx}{ax - c}\)= \(\frac{\lambda - 1}{\lambda + 1}\) are ...

If the roots of the equation x2−bxax−c\frac{x^{2} - bx}{ax - c}= λ−1λ+1\frac{\lambda - 1}{\lambda + 1} are such that

a + b = 0, then the value of l is-

A

a−ba+b\frac{a - b}{a + b}

B

c

C

1c\frac{1}{c}

D

a+ba−b\frac{a + b}{a - b}

Answer

a−ba+b\frac{a - b}{a + b}

Explanation

Solution

(l + 1) (x2 – bx) = (l – 1) (ax – c)

Ž lx2 + (–bl – b – al + a) x + cl – c = 0

Since a + b = 0, we have

(bλ−b−aλ+a)λ\frac{(b\lambda - b - a\lambda + a)}{\lambda} = 0

Ž –bl – b – al + a = 0 Ž l = a−ba+b\frac{a - b}{a + b}.