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

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

Smart Achievers · Accepted Answer

(4)

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

$∴$ $α + β = p, α β = r$

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

$∴$ $\frac{α}{2} + 2 β = q, α β = r$

Now, $α + β = p and α + 4 β = 2 q$

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

$= \frac{4 p - 2 q}{3} = \frac{2 (2 p - q)}{3}$

Hence, $r = α β = \frac{2}{9} (2 p - q) (2 q - p)$

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

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, 2βbe the roots of the equation x2-qx+r=0. Then the value of r is

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

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

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

4 29(2p-q)(2q-p)