If alpha and beta are roots of ax^2 + 2bx + c | vedclass

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

Question

(C) Given the quadratic equation $ax^2 + 2bx + c = 0$.From the properties of roots,the sum of roots $\alpha + \beta = -\frac{2b}{a}$ and the product o

Vedclass · Accepted Answer

$-\frac{2b}{\sqrt{ac}}$

Vedclass · Answer

(C) Given the quadratic equation $ax^2 + 2bx + c = 0$.From the properties of roots,the sum of roots $\alpha + \beta = -\frac{2b}{a}$ and the product of roots $\alpha \beta = \frac{c}{a}$.We need to evaluate the expression $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$.Taking the common denominator,we get $\frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}} = \frac{(\sqrt{\alpha})^2 + (\sqrt{\beta})^2}{\sqrt{\alpha \beta}} = \frac{\alpha + \beta}{\sqrt{\alpha \beta}}$.Substituting the values of $\alpha + \beta$ and $\alpha \beta$:$= \frac{-2b/a}{\sqrt{c/a}} = \frac{-2b/a}{\sqrt{c}/\sqrt{a}} = -\frac{2b}{a} 	imes \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{a} \cdot \sqrt{a}} 	imes \frac{\sqrt{a}}{\sqrt{c}} = -\frac{2b}{\sqrt{ac}}$.

If $\alpha$ and $\beta$ are roots of $ax^2 + 2bx + c = 0$,then $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$ is equal to

Similar Questions

If the sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is equal to the sum of the squares of their reciprocals,then $a/c, b/a, c/b$ are in

If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root and $a, b, c$ are non-zero real numbers,then $\frac{a^3 + b^3 + c^3}{abc} = $

If $a^{2}=by+cz, b^{2}=cz+ax, c^{2}=ax+by,$ then the value of $\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$ is

The roots of the equation $ax^2 + bx + c = 0$ will be reciprocal if

Given that $\tan \alpha$ and $\tan \beta$ are the roots of the equation $x^2 - px + q = 0$,then the value of $\sin^2(\alpha + \beta)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module