If alpha, beta and gamma are the roots of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2+4x+5=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -3$ $\alpha\beta+\beta\gamma+\gamma\al

Vedclass · Accepted Answer

$x^3+9x^2+43x+267=0$

Vedclass · Answer

(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2+4x+5=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -3$ $\alpha\beta+\beta\gamma+\gamma\alpha = 4$ $\alpha\beta\gamma = -5$ Let $A=1+4\alpha, B=1+4\beta, C=1+4\gamma$. Sum of roots: $A+B+C = 3+4(\alpha+\beta+\gamma) = 3+4(-3) = -9$. Sum of roots taken two at a time: $AB+BC+CA = (1+4\alpha)(1+4\beta) + (1+4\beta)(1+4\gamma) + (1+4\gamma)(1+4\alpha) = 3+8(\alpha+\beta+\gamma) + 16(\alpha\beta+\beta\gamma+\gamma\alpha) = 3+8(-3)+16(4) = 3-24+64 = 43$. Product of roots: $ABC = (1+4\alpha)(1+4\beta)(1+4\gamma) = 1+4(\alpha+\beta+\gamma)+16(\alpha\beta+\beta\gamma+\gamma\alpha)+64(\alpha\beta\gamma) = 1+4(-3)+16(4)+64(-5) = 1-12+64-320 = -267$. The required cubic equation is $x^3 - (A+B+C)x^2 + (AB+BC+CA)x - ABC = 0$. Substituting the values: $x^3 - (-9)x^2 + 43x - (-267) = 0$,which simplifies to $x^3+9x^2+43x+267=0$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+3x^2+4x+5=0$,then the cubic equation whose roots are $1+4\alpha$,$1+4\beta$ and $1+4\gamma$ is

Similar Questions

If the roots of the equation $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$ are equal in magnitude but opposite in sign,then $m = \dots$

If the product of the roots of the equation $x^2+4kx+12e^{3\log k}-1=0$ where $k>0$ is $323$,then the sum of its roots is:

Let $\alpha, \beta$ be the roots of the equation $x^2-px+r=0$ and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2-qx+r=0$. Then the value of $r$ is

If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + c = 0$,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-13x^2+kx+189=0$ such that $\beta-\gamma=2$,then $\beta+\gamma: k+\alpha=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module