Question

Let α, β be the roots of the equation x² - px + r = 0 and \(\dfrac{\alpha}{2}, 2\beta \) be the roots of the equation x² - qx + r = 0. Then the value of r is
1. \(\dfrac{2}{9}(p-q)(2q-p)\)
2. \(\dfrac{2}{9}(q-p)(2q-p)\)
3. \(\dfrac{2}{9}(q-2p)(2q-p)\)
4. \(\dfrac{2}{9}(2p-q)(2q-p)\)

Answer 1

Correct Answer - Option 1 : \(\dfrac{2}{9}(p-q)(2q-p)\)

Concept:

We know that, if α1, α2 be the roots of the quadratic equation ax2 + bx + c = 0 then 

α1 + α2 = \(\rm \dfrac {-b}{a}\) and α1 . α2 = \(\rm \dfrac {c}{a}\) 

Calculations:

We know that, if α1, α2 be the roots of the quadratic equation ax2 + bx + c = 0 then 

α1 + α2 = \(\rm \dfrac {-b}{a}\) and α1 . α2 = \(\rm \dfrac {c}{a}\) 

Given, α, β be the roots of the equation x2 - px + r = 0 

⇒αβ = r and α + β = p .... (1)

Also,  \(\dfrac{\alpha}{2}, 2\beta \) be the roots of the equation x2 - qx + r = 0

⇒\( \alpha\)β = r and \(\dfrac \alpha 2\) + 2β = q 

⇒ α + 4β = 2q .... (2)

subtracting equation (1) and (2), we get 

⇒ 3β = 2q - p 

⇒ β = \(\rm \dfrac {2q-p}{3}\)

Put the value of β in equation (1) , we get

⇒ α = p -  \(\rm \dfrac {2q-p}{3}\) 

⇒ α = \(\rm \dfrac {2p-2q}{3}\)

We have, r = αβ 

⇒r =  \(\rm \dfrac {2p-2q}{3}\) \(\rm \dfrac {2q-p}{3}\)

⇒r  = \(\dfrac{2}{9}(p-q)(2q-p)\)

Hence, if α, β be the roots of the equation x2 - px + r = 0 and \(\dfrac{\alpha}{2}, \beta \) be the roots of the equation x2 - qx + r = 0. Then the value of r is \(\dfrac{2}{9}(p-q)(2q-p)\)