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 ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of r

Vedclass · Accepted Answer

${n^2}$

Vedclass · Answer

(C) Given the quadratic equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha \beta = c/a$.Here,$\alpha + \beta = 1 + {n^2}$ and $\alpha \beta = \frac{1}{2}(1 + {n^2} + {n^4})$.We know that ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta$.Substituting the values:${\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 $x = \sqrt{7 + 4\sqrt{3}},$ then $x + \frac{1}{x} = $

The number of solutions of the equation $\log_{\sqrt{3}}(x^3 - 1) = \log_{\sqrt{3}}(x - 1) + 2$ is:

The roots of $|x - 2|^2 + |x - 2| - 6 = 0$ are

Let $S$ be the set of all real roots of the equation,$3^{x}(3^{x}-1)+2=|3^{x}-1|+|3^{x}-2|$. Then $S$

The values of $a$ and $b$ for which the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module