If alpha, beta are the roots of the equation | vedclass

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

Question

(B) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^{2}+b x+b=0$.From the properties of roots,we have the sum of roots $\

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^{2}+b x+b=0$.From the properties of roots,we have the sum of roots $\alpha+\beta = -\frac{b}{a}$ and the product of roots $\alpha \beta = \frac{b}{a}$.Now,we need to evaluate the expression $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{b}{a}}$.Simplifying the first two terms: $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}} = \frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}} = \frac{(\sqrt{\alpha})^2 + (\sqrt{\beta})^2}{\sqrt{\alpha \beta}} = \frac{\alpha+\beta}{\sqrt{\alpha \beta}}$.Substituting the values of $\alpha+\beta$ and $\alpha \beta$ into the expression:$= \frac{-\frac{b}{a}}{\sqrt{\frac{b}{a}}} + \sqrt{\frac{b}{a}}$.Since $\frac{x}{\sqrt{x}} = \sqrt{x}$,we get:$= -\sqrt{\frac{b}{a}} + \sqrt{\frac{b}{a}} = 0$.

If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+b=0,$ then $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{b}{a}}=$

Similar Questions

The number of real roots of the equation ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$ is:

If $x+\frac{1}{x}=5,$ then $x^{6}+\frac{1}{x^{6}}$ is

If $a + b + c = 0$,then the roots of the equation $4ax^2 + 3bx + 2c = 0$ are

The interval of $k$ for which the smaller root of the equation $x^2+kx-4=0$ lies in the interval $(-1, 2)$ is:

Sum of roots is $-1$ and sum of their reciprocals is $\frac{1}{6}$,then the equation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module