JEE Challenger
More from Complex Numbers and Quadratic Equations
MathMedAdv 2007 • Paper 1

Find Value of Constant r from Roots of Quadratic Equations

Let α,β\alpha, \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=0x^2 - qx + r = 0. Then the value of rr is

Options

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)

Correct

Topics & Concepts

Complex Numbers and Quadratic EquationsQuadratic Equations

Step-by-Step Solution

To find the value of rr, we use the relations between the roots and coefficients of the given quadratic equations.

For the equation x2−px+r=0x^2 - px + r = 0, the roots are α\alpha and β\beta. By Vieta's formulas:
α+β=p— (1)\alpha + \beta = p \quad \text{--- (1)}
αβ=r— (2)\alpha\beta = r \quad \text{--- (2)}

For the equation x2−qx+r=0x^2 - qx + r = 0, the roots are α2\frac{\alpha}{2} and 2β2\beta. By Vieta's formulas:
α2+2β=q— (3)\frac{\alpha}{2} + 2\beta = q \quad \text{--- (3)}
(α2)(2β)=αβ=r\left(\frac{\alpha}{2}\right)(2\beta) = \alpha\beta = r
which is consistent with equation (2).

Multiplying equation (3) by 22, we obtain:
α+4β=2q— (4)\alpha + 4\beta = 2q \quad \text{--- (4)}

Now, subtract equation (1) from equation (4):
(α+4β)−(α+β)=2q−p(\alpha + 4\beta) - (\alpha + \beta) = 2q - p
3β=2q−p  ⟹  β=2q−p33\beta = 2q - p \implies \beta = \frac{2q - p}{3}

Substitute the value of β\beta back into equation (1) to solve for α\alpha:
α=p−β=p−2q−p3=3p−(2q−p)3=4p−2q3=2(2p−q)3\alpha = p - \beta = p - \frac{2q - p}{3} = \frac{3p - (2q - p)}{3} = \frac{4p - 2q}{3} = \frac{2(2p - q)}{3}

Now, calculate r=αβr = \alpha\beta:
r=(2(2p−q)3)(2q−p3)=29(2p−q)(2q−p)r = \left(\frac{2(2p - q)}{3}\right) \left(\frac{2q - p}{3}\right) = \frac{2}{9}(2p - q)(2q - p)

Therefore, the correct option is D.

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