1.

If a, b are the real roots of `x^(2) + px + 1 = 0` and c, d are the real roots of `x^(2) + qx + 1 = 0`, then `(a-c)(b-c)(a+d)(b+d)` is divisible byA. `a - b - c - d`B. `a + b + c - d`C. `a + b + c + d`D. `a - b - c - d`

Answer» Correct Answer - C
We have, `a+b = -p, ab = 1, c+d = -q and cd = 1`.
Now,
`(a-c)(b-c)(a+d)(b+d)`
`=(a-c)(b+d)(b-c)(a+d)`
`=(ab+ad-bc-cd)(ba+bd-ca-cd)`
`=(ad-bc)(bd-ca)`
`=abd^(2)+abc^(2)-cda^(2)-b^(2)cd`
`=d^(2)+c^(2)-a^(2)-b^(2)`
`=(c+d)^(2) -(a+b)^(2)" "[because ab = cd = 1]`
`=(c+d+a+b)(c+d-a-b)`
`=-(a+b+c+d)(a+b-c-d)`
Hence, (a-c)(b-c)(a+d)(b+d) is divisible by a + b + c + d and a + b - c - d.


Discussion

No Comment Found

Related InterviewSolutions