If alpha and beta are the roots of the equati | vedclass

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

Question

(C) Given the quadratic equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$.From the relation between roots and coefficients,we have:

Vedclass · Accepted Answer

${n^2}$

Vedclass · Answer

(C) Given the quadratic equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$.From the relation between roots and coefficients,we have:$\alpha + \beta = 1 + {n^2}$$\alpha \beta = \frac{1}{2}(1 + {n^2} + {n^4})$We need to find the value of ${\alpha ^2} + {\beta ^2}$.Using the identity ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta$:${\alpha ^2} + {\beta ^2} = {(1 + {n^2})^2} - 2 	imes \frac{1}{2}(1 + {n^2} + {n^4})$${\alpha ^2} + {\beta ^2} = (1 + {n^4} + 2{n^2}) - (1 + {n^2} + {n^4})$${\alpha ^2} + {\beta ^2} = 1 + {n^4} + 2{n^2} - 1 - {n^2} - {n^4}$${\alpha ^2} + {\beta ^2} = {n^2}$.

If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$,then the value of ${\alpha ^2} + {\beta ^2}$ is

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the roots of ${x^3} + px + q = 0$,then the value of ${\alpha^3} + {\beta^3} + {\gamma^3}$ is equal to

If $\alpha, \beta, \gamma$ are roots of the equation $x^3 + ax^2 + bx + c = 0$,then $\alpha^{-1} + \beta^{-1} + \gamma^{-1} = $

If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation $x^2 + bx + c = 0,$ where $c < 0 < b,$ then

If the harmonic mean of the roots of the equation $\sqrt{2} x^2 - bx + (8 - 2\sqrt{5}) = 0$ is $4$,then the value of $b$ is

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+3x-4=0$,then the value of $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module