If the roots of the equation x^2+px+q=0 are a | 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 $\alpha, \beta$,we have $\alpha+\beta = -p$ and $\alpha\beta = q$. Given the roots of $x^2-rx+s=0$ are $\alpha

Vedclass · Accepted Answer

Both real

Vedclass · Answer

(C) Given the roots of $x^2+px+q=0$ are $\alpha, \beta$,we have $\alpha+\beta = -p$ and $\alpha\beta = q$. Given the roots of $x^2-rx+s=0$ are $\alpha^4, \beta^4$,we have $\alpha^4+\beta^4 = r$ and $\alpha^4\beta^4 = s$. Consider the quadratic 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 $r = \alpha^4+\beta^4$,we get $D = 8(\alpha\beta)^2 + 4(\alpha^4+\beta^4) = 4(2\alpha^2\beta^2 + \alpha^4+\beta^4) = 4(\alpha^2+\beta^2)^2$. Since $(\alpha^2+\beta^2)^2 \geq 0$,it follows that $D \geq 0$. Therefore,the roots of the equation are always real.

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

Similar Questions

The roots of the equation $x^3-3x-2=0$ are

Water rises to a height $h$ in a capillary tube at the surface of the Earth. On the surface of the Moon,the height of the water column in the same capillary tube will be:

When a sound wave of wavelength $\lambda$ is propagating in a medium,the maximum velocity of the particle is equal to the wave velocity. The amplitude of the wave is:

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides of the triangle are in

The foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ coincide. Then the value of $b^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module