Question

If `x^2+(a-b)x=(1-a-b)=0. w h e r ea ,b in R ,` then find the values of `a` for which equation has unequal real roots for all values of `bdot`

Answer 1

Correct Answer - ` a gt 1`
The given equation is , ` x^(2) + (a - b) x + (1 - a - b) = 0`
` a, b in `R
For this equations to have unequal real roots for all b, D `gt` 0
`rAr (a - b)^(2) - 4 (1 - a - b) gt 0`
`rArr a^(2) + b^(2) - 2ab - 4 + 4 a + 4b gt 0 `
`rArr b^(2) + b (4 - 2a) + a^(2) + 4a - 4 gt 0` (1)
This is a quadratic expression in b, and it must be true `AA be in ` R
So, Discriminant `lt 0`
`rArr (4 -2 a)^(2) - 4 a - 4) lt 0`
`rArr (2 - a)^(2) - (a^(2) + 4a - 4) lt 0 `
`rArr 4 -4a + a^(2) - a^(2) - 4a + 4 lt 0`
` rArr - 8a + 8 lt 0 ` n
`rArr a gt 1` .