If alpha and beta are the roots of x^2+7x+3=0 | vedclass

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(D) Let $y = \frac{2\alpha}{3-4\alpha}$. Then $2\alpha = 3y - 4\alpha y$,which implies $\alpha(2+4y) = 3y$,so $\alpha = \frac{3y}{2+4y}$.Since $\alpha

Vedclass · Accepted Answer

$81$

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(D) Let $y = \frac{2\alpha}{3-4\alpha}$. Then $2\alpha = 3y - 4\alpha y$,which implies $\alpha(2+4y) = 3y$,so $\alpha = \frac{3y}{2+4y}$.Since $\alpha$ is a root of $x^2+7x+3=0$,we substitute $\alpha$:$(\frac{3y}{2+4y})^2 + 7(\frac{3y}{2+4y}) + 3 = 0$.Multiplying by $(2+4y)^2 = 16y^2+16y+4$:$9y^2 + 21y(2+4y) + 3(16y^2+16y+4) = 0$.$9y^2 + 42y + 84y^2 + 48y^2 + 48y + 12 = 0$.$141y^2 + 90y + 12 = 0$.Dividing by $3$: $47y^2 + 30y + 4 = 0$.Comparing with $ax^2+bx+c=0$,we get $a=47, b=30, c=4$. Since $GCD(47, 30, 4) = 1$,the values are valid.Thus,$a+b+c = 47+30+4 = 81$.

If $\alpha$ and $\beta$ are the roots of $x^2+7x+3=0$ and $\frac{2\alpha}{3-4\alpha}, \frac{2\beta}{3-4\beta}$ are the roots of $ax^2+bx+c=0$ and $GCD(a, b, c) = 1$,then $a+b+c=$

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