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If a x^2+(b-c)x+a-b-c=0 has unequal real...

If `a x^2+(b-c)x+a-b-c=0` has unequal real roots for all `c in R ,t h e n` `b<0 a >0`

A

` b lt 0 lt a `

B

`a lt 0 lt b `

C

`b lt alt 0 `

D

`b gt a gt 0 `

Text Solution

Verified by Experts

The correct Answer is:
3,4

We have,
` D = (b - c)^(2) - 4a(a - b - c) gt 0 `
or ` b^(2) + c^(2) - 2bc - 4a^(2) + 4ab + 4ac gt 0 `
` or c^(2) + (4a - 2b) c - 4a^(2) + 4ab + b^(2) gt 0 ` for all ` c in ` R
Discriminant of the above expression in c must be negative
Hence,
` 1(4a = 2b)^(2) - 4 (-4a^(2) + 4ab + b^(2)) lt 0 `
or ` 4a^(2) - 4ab + b ^(2) + 4a^(2) - 4ab - b^(2) lt 0 `
or ` a (a - b) lt 0 `
`rArr a lt 0 and a - b gt 0 or a gt 0 and a - b lt 0 `
`rArr b lt a lt 0 or b gt a gt 0 ` .
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