Question
Let $\alpha,\beta$ be the roots of the equation $x^{2} - 3x + r = 0$, and $\frac{\alpha}{2},2\beta$ be the roots of the equation $x^{2} + 3x + r = 0$. If the roots of the equation $x^{2} + 6x = m$ are $2\alpha + \beta + 2r$ and $\alpha - 2\beta - \frac{r}{2}$, then m is equal to :-
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Solution
Sol. [Key Concept: Use sum & product of roots for both quadratics to find $\alpha,\beta,r$; then $m$ from the third equation.]
$$\alpha + \beta = 3$$
$$\frac{\alpha}{2} + 2\beta = - 3$$
On solving, we get $\alpha = 6;\beta = - 3$
Product of roots $= \alpha\beta = r \Rightarrow r = - 18$
Now for $x^{2} + 6x - m = 0$
Product of roots $= - m$
$${= (2\alpha + \beta + 2r)\left( \alpha - 2\beta - \frac{r}{2} \right) }{\Rightarrow - m = ( - 27)(21) }{\Rightarrow m = 567 }$$
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