If the roots of the equation x^2 + px + q = 0 | vedclass

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

Question

(C) Given the roots of $x^2 + px + q = 0$ are $\alpha$ and $\beta$,we have $\alpha + \beta = -p$ and $\alpha\beta = q$.Given the roots of $x^2 - xr +

Vedclass · Accepted Answer

Both real

Vedclass · Answer

(C) Given the roots of $x^2 + px + q = 0$ are $\alpha$ and $\beta$,we have $\alpha + \beta = -p$ and $\alpha\beta = q$.Given the roots of $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,we have $\alpha^4 + \beta^4 = r$ and $\alpha^4\beta^4 = s$.Consider the equation $x^2 - 4qx + 2q^2 - r = 0$. The discriminant $D$ is given by:$D = (-4q)^2 - 4(1)(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$.Substituting $q = \alpha\beta$ and $r = \alpha^4 + \beta^4$:$D = 8(\alpha\beta)^2 + 4(\alpha^4 + \beta^4) = 4(2\alpha^2\beta^2 + \alpha^4 + \beta^4) = 4(\alpha^2 + \beta^2)^2$.Since $(\alpha^2 + \beta^2)^2 \ge 0$,it follows that $D \ge 0$.Therefore,the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ are always real.

If the roots of the equation $x^2 + px + q = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,then the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ will be

Similar Questions

The real roots of the equation $x^2 + 5|x| + 4 = 0$ are

Let $p, q \in \mathbb{Q}$. If $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$,then:

The number of solutions of the equations $x+y+z=1$,$x^2+y^2+z^2=1$,and $x^3+y^3+z^3=1$ is:

The roots of the equation $x^2 - \sqrt{13}x + 1 = 0$ are:

The number of real numbers $x$ such that there exists an isosceles triangle having two of its angles measured in degrees equal to $2x + 7$ and $7x + 10$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module