The solution of the equation 2x^3 - x^2 - 22x | vedclass

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Question

(B) Given the cubic equation: $2x^3 - x^2 - 22x - 24 = 0$. Let the roots be $\alpha, \beta, \gamma$. Given the ratio of two roots is $3:4$,let the roo

Vedclass · Accepted Answer

$\frac{-3}{2}, -2, 4$

Vedclass · Answer

(B) Given the cubic equation: $2x^3 - x^2 - 22x - 24 = 0$. Let the roots be $\alpha, \beta, \gamma$. Given the ratio of two roots is $3:4$,let the roots be $3k, 4k, \gamma$. From the sum of roots: $3k + 4k + \gamma = -(\frac{-1}{2}) = \frac{1}{2}$ $\Rightarrow 7k + \gamma = \frac{1}{2}$ $\Rightarrow \gamma = \frac{1}{2} - 7k$. From the product of roots: $(3k)(4k)(\gamma) = -(\frac{-24}{2}) = 12$ $\Rightarrow 12k^2 \gamma = 12$ $\Rightarrow k^2 \gamma = 1$. Substituting $\gamma$: $k^2(\frac{1}{2} - 7k) = 1$ $\Rightarrow \frac{1}{2}k^2 - 7k^3 = 1$ $\Rightarrow 14k^3 - k^2 + 2 = 0$. Testing $k = -1/2$: $14(-1/8) - (1/4) + 2 = -7/4 - 1/4 + 2 = -2 + 2 = 0$. Thus,$k = -1/2$. The roots are $3(-1/2) = -3/2$,$4(-1/2) = -2$,and $\gamma = 1/(-1/2)^2 = 4$. The roots are $\frac{-3}{2}, -2, 4$.

The solution of the equation $2x^3 - x^2 - 22x - 24 = 0$,given that two of the roots are in the ratio $3:4$,is:

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