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IIT-JEE 2007 · Paper 1, Mathematics question 45

Let α,β\alpha,\beta be the roots of the equation x2−px+r=0x^2-px+r=0 and α2\frac\alpha2, 2β2\beta be the roots of the equation x2−qx+r=0x^2-qx+r=0. Then the value of rr is

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Answer and worked solution

Official answer: D

Vieta. α+β=p\alpha+\beta=p and α2+2β=q\frac\alpha2+2\beta=q (both products equal rr automatically, since α2⋅2β=αβ\frac\alpha2\cdot2\beta=\alpha\beta).

Solve. Doubling the second, α+4β=2q\alpha+4\beta=2q; subtracting the first, 3β=2q−p3\beta=2q-p. So
β=2q−p3,α=p−β=2(2p−q)3.\beta=\frac{2q-p}3,\qquad\alpha=p-\beta=\frac{2(2p-q)}3.

Product. r=αβ=29(2p−q)(2q−p)r=\alpha\beta=\dfrac29(2p-q)(2q-p).

(D)

Topics:Roots and coefficients
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