The two roots of the equation x^3 - 9x^2 + 14 | vedclass

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Question

(A) Let the roots be $3\alpha, 2\alpha, \beta$. According to Vieta's formulas for the cubic equation $x^3 - 9x^2 + 14x + 24 = 0$: $1) 	ext{Sum of roo

Vedclass · Accepted Answer

$6, 4, -1$

Vedclass · Answer

(A) Let the roots be $3\alpha, 2\alpha, \beta$. According to Vieta's formulas for the cubic equation $x^3 - 9x^2 + 14x + 24 = 0$: $1) 	ext{Sum of roots: } 3\alpha + 2\alpha + \beta = 9 \implies 5\alpha + \beta = 9 \implies \beta = 9 - 5\alpha$ $2) 	ext{Sum of product of roots taken two at a time: } (3\alpha)(2\alpha) + (2\alpha)(\beta) + (3\alpha)(\beta) = 14 \implies 6\alpha^2 + 5\alpha\beta = 14$ $3) 	ext{Product of roots: } (3\alpha)(2\alpha)(\beta) = -24 \implies 6\alpha^2\beta = -24 \implies \alpha^2\beta = -4$ Substitute $\beta = 9 - 5\alpha$ into the second equation: $6\alpha^2 + 5\alpha(9 - 5\alpha) = 14$ $6\alpha^2 + 45\alpha - 25\alpha^2 = 14$ $-19\alpha^2 + 45\alpha - 14 = 0 \implies 19\alpha^2 - 45\alpha + 14 = 0$ Factoring the quadratic: $(19\alpha - 7)(\alpha - 2) = 0$ So,$\alpha = 2$ or $\alpha = \frac{7}{19}$. If $\alpha = 2$,then $\beta = 9 - 5(2) = -1$. Check with the product equation: $\alpha^2\beta = (2)^2(-1) = -4$,which satisfies the condition. The roots are $3(2), 2(2), -1$,which are $6, 4, -1$.

The two roots of the equation $x^3 - 9x^2 + 14x + 24 = 0$ are in the ratio $3 : 2$. Find the roots.

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