If one root of x^3-7x^2+36=0 is twice the oth | vedclass

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(B) Let the roots of the cubic equation $x^3-7x^2+0x+36=0$ be $\alpha, 2\alpha,$ and $\beta$.From the relation between roots and coefficients:Sum of r

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Let the roots of the cubic equation $x^3-7x^2+0x+36=0$ be $\alpha, 2\alpha,$ and $\beta$.From the relation between roots and coefficients:Sum of roots: $\alpha + 2\alpha + \beta = 7 \implies 3\alpha + \beta = 7 \implies \beta = 7 - 3\alpha$.Sum of roots taken two at a time: $\alpha(2\alpha) + 2\alpha\beta + \beta\alpha = 0 \implies 2\alpha^2 + 3\alpha\beta = 0$.Since $\alpha 
eq 0$ (as $36 
eq 0$),we divide by $\alpha$: $2\alpha + 3\beta = 0$.Substitute $\beta = 7 - 3\alpha$ into the equation: $2\alpha + 3(7 - 3\alpha) = 0$.$2\alpha + 21 - 9\alpha = 0 \implies -7\alpha = -21 \implies \alpha = 3$.Then the roots are $\alpha = 3$ and $2\alpha = 6$.The sum of these two roots is $3 + 6 = 9$.

If one root of $x^3-7x^2+36=0$ is twice the other,then the sum of those two roots is

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