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Mathematics Question on Complex Numbers and Quadratic Equations

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

A

29(p−q)(2q−p)\frac{2}{9}(p-q)(2q-p)

B

29(q−p)(2p−q)\frac{2}{9}(q-p)(2p-q)

C

29(q−2p)(2q−p)\frac{2}{9}(q-2p)(2q-p)

D

29(2p−q)(2q−p)\frac{2}{9}(2p-q)(2q-p)

Answer

29(2p−q)(2q−p)\frac{2}{9}(2p-q)(2q-p)

Explanation

Solution

The equation x2−px+r=0x^2-px+r=0 has roots α\alpha, β\beta and the equation x2−qx+r=0x^2-qx+r=0 has roots α2,2β.\frac{\alpha}{2},2\beta.
⇒       r=αβ\Rightarrow\, \, \, \, \, \, \, r=\alpha\beta and α+β=p, \alpha+\beta=p,
and α2+2β=q⇒β=2q−p3\frac{\alpha}{2}+2\beta=q\Rightarrow\beta=\frac{2q-p}{3} and α=2(2p−q)3 \alpha=\frac{2(2p-q)}{3}
⇒       αβ=r=29(2q−p)(2p−q)\Rightarrow\, \, \, \, \, \, \, \alpha\beta=r=\frac{2}{9}(2q-p)(2p-q)