If the sum of the roots of the equation {x^2} | vedclass

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

Question

(C) Let the roots of the quadratic equation ${x^2} + px + q = 0$ be $\alpha$ and $\beta$.From the properties of quadratic equations,the sum of the roo

Vedclass · Accepted Answer

${p^2} + p = 2q$

Vedclass · Answer

(C) Let the roots of the quadratic equation ${x^2} + px + q = 0$ be $\alpha$ and $\beta$.From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = -p$ and the product of the roots is $\alpha \beta = q$.According to the problem,the sum of the roots is equal to the sum of their squares:$\alpha + \beta = {\alpha ^2} + {\beta ^2}$We know that ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta$.Substituting the values of the sum and product of the roots into the equation:$-p = {(-p)^2} - 2q$$-p = {p^2} - 2q$Rearranging the terms,we get:${p^2} + p = 2q$.

If the sum of the roots of the equation ${x^2} + px + q = 0$ is equal to the sum of their squares,then

Similar Questions

Solve the given two equations and select the correct option.
$I.$ $x^{2}-5x+6=0$
$II.$ $2y^{2}-15y+27=0$

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,then the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ is

Product of real roots of the equation ${t^2}{x^2} + |x| + 9 = 0$.

If $\alpha$ and $\beta$ are roots of the equation $Ax^2 + Bx + C = 0$,then the value of $\alpha^3 + \beta^3$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2} - 3481 = 0$
$II.$ $3y^{2} = \sqrt[3]{216000}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module