If the equation x^{2}-cx+d=0 has roots equal | vedclass

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

Question

(D) Let $\alpha$ and $\beta$ be the roots of $x^{2}+ax+b=0$ and the roots of $x^{2}-cx+d=0$ be $\alpha^{4}$ and $\beta^{4}$.
From the relations betwe

Vedclass · Accepted Answer

one positive and one negative

Vedclass · Answer

(D) Let $\alpha$ and $\beta$ be the roots of $x^{2}+ax+b=0$ and the roots of $x^{2}-cx+d=0$ be $\alpha^{4}$ and $\beta^{4}$.
From the relations between roots and coefficients:
$\alpha+\beta=-a, \alpha\beta=b$ ...$(i)$
$\alpha^{4}+\beta^{4}=c, \alpha^{4}\beta^{4}=d$ ...$(ii)$
From $(ii),$ $c = \alpha^{4}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2}-2(\alpha\beta)^{2} = ((\alpha+\beta)^{2}-2\alpha\beta)^{2}-2(\alpha\beta)^{2}$.
Substituting $(i)$: $c = (a^{2}-2b)^{2}-2b^{2} = a^{4}+4b^{2}-4a^{2}b-2b^{2} = a^{4}-4a^{2}b+2b^{2}$.
Thus,$2b^{2}-c = 2b^{2}-(a^{4}-4a^{2}b+2b^{2}) = 4a^{2}b-a^{4} = a^{2}(4b-a^{2})$.
Given $a^{2}>4b$,we have $4b-a^{2} < 0$,so $2b^{2}-c < 0$.
For the equation $x^{2}-4bx+(2b^{2}-c)=0$,the product of roots is $2b^{2}-c < 0$.
Since the product of the roots is negative,one root must be positive and the other must be negative.

If the equation $x^{2}-cx+d=0$ has roots equal to the fourth powers of the roots of $x^{2}+ax+b=0,$ where $a^{2}>4b,$ then the roots of $x^{2}-4bx+2b^{2}-c=0$ will be

Similar Questions

If the roots of the quadratic equation $x^2-35x+c=0$ are in the ratio $2:3$ and $c=6K$,then $K=$

If the sum of the roots of a quadratic equation is $1$ and the sum of the squares of the roots is $13$,then find the equation.

If $\alpha, \beta, \gamma$ are the roots of $x^3+2x^2-3x-1=0$,then $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=$

What is the geometric mean of the roots of the equation $x^2 - 18x + 9 = 0$?

If the harmonic mean between the roots of $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$ is $4$,then the value of $b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module