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If the zeroes of the polynomial $x^2 + ax + b$ are in the ratio $3: 4$, then prove that $12a^2 = 49b$.
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Let the zeroes are $3\alpha$ and $4\alpha$
$3\alpha + 4\alpha = -a$
$\Rightarrow 7\alpha = -a$ (1/2)
Also, $12\alpha^2 = b$ (1/2)
LHS = $12a^2 = 12 (-7\alpha)^2 = 49 \times 12(\alpha)^2 = 49b$ = RHS (1)
$3\alpha + 4\alpha = -a$
$\Rightarrow 7\alpha = -a$ (1/2)
Also, $12\alpha^2 = b$ (1/2)
LHS = $12a^2 = 12 (-7\alpha)^2 = 49 \times 12(\alpha)^2 = 49b$ = RHS (1)