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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` |
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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. |
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