If `alpha,beta` are the roots of `ax^2 + c = bx`, then the equation `(a + cy)^2 =b^2y` in y has the roots

Options

A. `alpha beta^(-1), alpha^(-1)beta`
B. `alpha^(-2), beta_(-2)`
C. `alpha^(-1), beta^(-1)`
D. `alpha^(2), beta^(2)`

Answer

2

Explanation

`ax^(2) - bx + c = 0`
`alpha + beta = b/a, abeta = c/a`
Also, `(a + cy)^(2) = b^(2)y`
`rArr c^(2)y^(2) - (b^(2) - 2ac)y + a^(2) = 0`
`rArr (c/a)^(2)y^(2) - ((b/a)^(2) - 2(c/a))y + 1 = 0`
`rArr (alphabeta)^(2)y^(2) - (alpha^(2) + beta^(2))y + 1 = 0`
`rArr y^(2) - (alpha^(-2) + beta^(-2))y + alpha^(-2)beta^(-2) = 0`
`rArr (y - alpha^(-2)) (y - beta^(-2)) = 0`
Hence, the roots are `alpha^(-2), beta^(-2)`.