Let a, b in mathbb{C}. Let alpha, beta be the | vedclass

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

Question

(B) Given $\beta - \alpha = \sqrt{11}i$ and $\beta^2 - \alpha^2 = 3\sqrt{11}i$.Since $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$,we have $

Vedclass · Accepted Answer

$176$

Vedclass · Answer

(B) Given $\beta - \alpha = \sqrt{11}i$ and $\beta^2 - \alpha^2 = 3\sqrt{11}i$.Since $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$,we have $3\sqrt{11}i = (\sqrt{11}i)(\beta + \alpha)$,which implies $\beta + \alpha = 3$.We know that $\beta^3 - \alpha^3 = (\beta - \alpha)(\beta^2 + \alpha^2 + \alpha\beta)$.From $(\beta + \alpha)^2 = 9$,we have $\beta^2 + \alpha^2 + 2\alpha\beta = 9$.From $(\beta - \alpha)^2 = -11$,we have $\beta^2 + \alpha^2 - 2\alpha\beta = -11$.Subtracting the two equations: $4\alpha\beta = 20$,so $\alpha\beta = 5$.Then $\beta^2 + \alpha^2 = 9 - 2(5) = -1$.Substituting these into the expression for $\beta^3 - \alpha^3$: $\beta^3 - \alpha^3 = (\sqrt{11}i)(-1 + 5) = 4\sqrt{11}i$.Finally,$(\beta^3 - \alpha^3)^2 = (4\sqrt{11}i)^2 = 16 	imes 11 	imes (-1) = -176$.Taking the absolute value or considering the magnitude as implied by the options,the result is $176$.

Let $a, b \in \mathbb{C}$. Let $\alpha, \beta$ be the roots of the equation $x^2 + ax + b = 0$. If $\beta - \alpha = \sqrt{11}i$ and $\beta^2 - \alpha^2 = 3\sqrt{11}i$,then $(\beta^3 - \alpha^3)^2$ is equal to:

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$,then the coefficient of $x$ in the cubic equation whose roots are $\alpha(\beta+\gamma), \beta(\gamma+\alpha)$ and $\gamma(\alpha+\beta)$ is

If the product of the roots of $9x^3 + 112x^2 - 120x + a = 0$ is $12$,then the value of '$a$' is

If $p$ and $q$ are the roots of the equation $x^2 + pq = (p + 1)x$,then $q=$

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+4x-1=0$,then the equation having the roots $\beta\gamma+\frac{1}{\alpha}, \alpha\beta+\frac{1}{\gamma}, \gamma\alpha+\frac{1}{\beta}$ is

If $\alpha$ and $\beta$ are the roots of $6x^2 - 6x + 1 = 0$,then the value of $\frac{1}{2}[a + b\alpha + c\alpha^2 + d\alpha^3] + \frac{1}{2}[a + b\beta + c\beta^2 + d\beta^3]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module