Let be the roots of the equation x 2 - p x + r = 0 and 2 , 2 be the roots of the equation x 2 - q x + r = 0 . Then the value of r is [2007]

Question

Let $α, β$ be the roots of the equation $x^{2} - p x + r = 0$ and $\frac{α}{2}, 2 β$ be the roots of the equation $x^{2} - q x + r = 0$ . Then the value of $r$ is [2007]

Smart Achievers · Accepted Answer

(4)

$Since α and β are the roots of x^{2} - p x + r = 0$

$∴$ $α + β = p \dots (i)$

$r$

$Also \frac{α}{2} and 2 β are the roots of x^{2} - q x + r = 0$

$∴$ $\frac{α}{2} + 2 β = q \Rightarrow α + 4 β = 2 q \dots (i i i)$

$Solving (i) and (iii) for α and β, we get$

$β = \frac{1}{3} (2 q - p) and α = \frac{2}{3} (2 p - q)$

$On substituting the values of α and β in equation (ii),$

$we get \frac{2}{9} (2 p - q) (2 q - p) = r .$

Solution

Q.
Let $α, β$ be the roots of the equation $x^{2} - p x + r = 0$ and $\frac{α}{2}, 2 β$ be the roots of the equation $x^{2} - q x + r = 0$ . Then the value of $r$ is [2007]

1 $\frac{2}{9} (p - q) (2 q - p)$

2 $\frac{2}{9} (q - p) (2 p - q)$

3 $\frac{2}{9} (q - 2 p) (2 q - p)$

4 $\frac{2}{9} (2 p - q) (2 q - p)$

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Solution

Q. Let α,β be the roots of the equation x2-px+r=0 and α2,2β be the roots of the equation x2-qx+r=0. Then the value of r is [2007]

1 29(p-q)(2q-p)

2 29(q-p)(2p-q)

3 29(q-2p)(2q-p)