If alpha, beta are the roots of x^2-3x+a=0 an | vedclass

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(C) Let the roots be $\alpha, \alpha r, \alpha r^2, \alpha r^3$ where $r > 1$. From the first equation $x^2-3x+a=0$,we have $\alpha + \alpha r = 3$ an

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(C) Let the roots be $\alpha, \alpha r, \alpha r^2, \alpha r^3$ where $r > 1$. From the first equation $x^2-3x+a=0$,we have $\alpha + \alpha r = 3$ and $\alpha(\alpha r) = a$. From the second equation $x^2-12x+b=0$,we have $\alpha r^2 + \alpha r^3 = 12$ and $(\alpha r^2)(\alpha r^3) = b$. From $\alpha(1+r) = 3$ and $\alpha r^2(1+r) = 12$,we divide the two equations: $\frac{\alpha r^2(1+r)}{\alpha(1+r)} = \frac{12}{3}$,which gives $r^2 = 4$. Since $r > 1$,we have $r = 2$. Substituting $r=2$ into $\alpha(1+r) = 3$,we get $\alpha(3) = 3$,so $\alpha = 1$. The roots are $1, 2, 4, 8$. Thus,$a = \alpha(\alpha r) = 1 	imes 2 = 2$ and $b = (\alpha r^2)(\alpha r^3) = 4 	imes 8 = 32$. Therefore,$a+b = 2+32 = 34$.

If $\alpha, \beta$ are the roots of $x^2-3x+a=0$ and $\gamma, \delta$ are the roots of $x^2-12x+b=0$ and $\alpha, \beta, \gamma, \delta$ in that order form a geometric progression in increasing order with common ratio $r>1$,then $a+b=$

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