If the sum of the roots of the quadratic equa | vedclass

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

Question

(A) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.From the properties of roots,we have $\alpha + \beta = -\frac{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.From the properties of roots,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.According to the problem,the sum of the roots is equal to the sum of the squares of their reciprocals:$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$$\alpha + \beta = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$Substitute $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:$-\frac{b}{a} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$$-\frac{b}{a} = \frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2}$$-\frac{b}{a} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}}$Multiply both sides by $\frac{c^2}{a^2}$:$-\frac{b}{a} \cdot \frac{c^2}{a^2} = \frac{b^2}{a^2} - \frac{2c}{a}$$-\frac{bc^2}{a^3} = \frac{b^2}{a^2} - \frac{2c}{a}$Divide the entire equation by $\frac{c}{a}$ (assuming $c 
eq 0$):$-\frac{bc}{a^2} = \frac{b^2}{ac} - 2$Rearranging the terms,we get:$\frac{b^2}{ac} + \frac{bc}{a^2} = 2$.

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 $\frac{b^2}{ac} + \frac{bc}{a^2} = $

Similar Questions

With respect to the roots of $x^{2}-x-2=0$,we can say that

If the roots of the equation $Ax^2 + Bx + C = 0$ are $\alpha, \beta$ and the roots of the equation $x^2 + px + q = 0$ are $\alpha^2, \beta^2$,then the value of $p$ will be

If $\lambda \in R$ is such that the sum of the cubes of the roots of the equation $x^2 + (2 - \lambda) x + (10 - \lambda) = 0$ is minimum,then the magnitude of the difference of the roots of this equation 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

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt{2304}$
$II.$ $y^2 = 2304$

Vedclass Test Series

Exam Paper Generator

Online Exam Module