Let , be the roots of the equation , and , be the roots of the equation . If the roots of the equation are and , then m is equal to [2026]

Question

Let $α$ , $β$ be the roots of the equation $x^{2} - 3 x + r = 0$ , and $\frac{α}{2}$ , $2 β$ be the roots of the equation $x^{2} + 3 x + r = 0$ . If the roots of the equation $x^{2} + 6 x = m$ are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2}$ , then m is equal to [2026]

Smart Achievers · Accepted Answer

(4)

$α$ , $β$ are the roots of $x^{2} - 3 x + r = 0$

$∴ α + β = 3 and α β = r$

$\frac{α}{2}, 2 β$ are the roots of $x^{2} + 3 x + r = 0$

$∴ \frac{α}{2} + 2 β = - 3 and α β = r$

Now, $α + β = 3 and α + 4 β = - 6$

$\Rightarrow β = - 3, α = 6 and r = - 18$

Now, roots are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2} = 12 - 3 - 36, 6 + 6 + 9 = - 27, 21$

$∴$ Roots of $x^{2} + 6 x - m = 0$ are –27, 21

$∴$ (–27)(21) = –m $\Rightarrow$ m = (27)(21) $\Rightarrow$ m = 567

Solution

Q.
Let $α$ , $β$ be the roots of the equation $x^{2} - 3 x + r = 0$ , and $\frac{α}{2}$ , $2 β$ be the roots of the equation $x^{2} + 3 x + r = 0$ . If the roots of the equation $x^{2} + 6 x = m$ are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2}$ , then m is equal to [2026]

1 –135

2 –567

3 135

4 567

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