If each root of the equation 2x^3 + ax^2 - 8x | vedclass

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Question

(B) Let the roots of the equation $2x^3 + ax^2 - 8x + b = 0$ be $\alpha, \beta, \gamma$. After reducing each root by $1$,the new roots are $\alpha-1,

Vedclass · Accepted Answer

$1, 1 \pm \sqrt{7}$

Vedclass · Answer

(B) Let the roots of the equation $2x^3 + ax^2 - 8x + b = 0$ be $\alpha, \beta, \gamma$. After reducing each root by $1$,the new roots are $\alpha-1, \beta-1, \gamma-1$. Let $y = x - 1$,so $x = y + 1$. Substituting $x = y + 1$ into the original equation: $2(y+1)^3 + a(y+1)^2 - 8(y+1) + b = 0$ $2(y^3 + 3y^2 + 3y + 1) + a(y^2 + 2y + 1) - 8(y + 1) + b = 0$ $2y^3 + (6 + a)y^2 + (6 + 2a - 8)y + (2 + a - 8 + b) = 0$. Given that the coefficient of $y^2$ and the constant term vanish: $6 + a = 0 \Rightarrow a = -6$. $2 + a - 8 + b = 0$ $\Rightarrow 2 - 6 - 8 + b = 0$ $\Rightarrow b = 12$. The original equation is $2x^3 - 6x^2 - 8x + 12 = 0$. Dividing by $2$,we get $x^3 - 3x^2 - 4x + 6 = 0$. Since $x = 1$ is a root $(1 - 3 - 4 + 6 = 0)$,we divide by $(x - 1)$: $(x - 1)(x^2 - 2x - 6) = 0$. The roots are $x = 1$ and $x = \frac{2 \pm \sqrt{4 - 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$. Thus,the roots are $1, 1 + \sqrt{7}, 1 - \sqrt{7}$.

If each root of the equation $2x^3 + ax^2 - 8x + b = 0$ is reduced by $1$,then in the transformed equation thus formed,the term containing $x^2$ and the constant term vanish. The roots of the original equation are

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