Let x_{1}, x_{2} be the roots of x^{2}-3x+a=0 | vedclass

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(B) Given that $x_{1}, x_{2}$ are roots of $x^{2}-3x+a=0$,so $x_{1}+x_{2}=3$ and $x_{1}x_{2}=a$. Given that $x_{3}, x_{4}$ are roots of $x^{2}-12x+b=0

Vedclass · Accepted Answer

$64$

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(B) Given that $x_{1}, x_{2}$ are roots of $x^{2}-3x+a=0$,so $x_{1}+x_{2}=3$ and $x_{1}x_{2}=a$. Given that $x_{3}, x_{4}$ are roots of $x^{2}-12x+b=0$,so $x_{3}+x_{4}=12$ and $x_{3}x_{4}=b$. Since $x_{1}, x_{2}, x_{3}, x_{4}$ are in $GP$,let them be $A, AR, AR^{2}, AR^{3}$. Then $x_{1}+x_{2} = A(1+R) = 3$ and $x_{3}+x_{4} = AR^{2}(1+R) = 12$. Dividing the two equations,we get $R^{2} = 12/3 = 4$,so $R = 2$ (since $x_{1}  0$). Substituting $R=2$ into $A(1+R)=3$,we get $A(3)=3$,so $A=1$. The terms are $1, 2, 4, 8$. Thus,$a = x_{1}x_{2} = 1 	imes 2 = 2$ and $b = x_{3}x_{4} = 4 	imes 8 = 32$. Therefore,$ab = 2 	imes 32 = 64$.

Let $x_{1}, x_{2}$ be the roots of $x^{2}-3x+a=0$ and $x_{3}, x_{4}$ be the roots of $x^{2}-12x+b=0$. If $x_{1} < x_{2} < x_{3} < x_{4}$ and $x_{1}, x_{2}, x_{3}, x_{4}$ are in $GP$,then $ab$ equals:

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