If alpha_1, beta_1, gamma_1, delta_1 are the | vedclass

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

Question

(D) Given that $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of $a x^4+b x^3+c x^2+d x+e=0$. The roots of $e x^4+d x^3+c x^2+b x+a=0$ are the

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given that $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of $a x^4+b x^3+c x^2+d x+e=0$. The roots of $e x^4+d x^3+c x^2+b x+a=0$ are the reciprocals of the roots of the first equation. Thus,$\alpha_2 = \frac{1}{\delta_1}, \beta_2 = \frac{1}{\gamma_1}, \gamma_2 = \frac{1}{\beta_1}, \delta_2 = \frac{1}{\alpha_1}$. Given $\alpha_1 - \delta_2 = 2 \implies \alpha_1 - \frac{1}{\alpha_1} = 2 \implies \alpha_1^2 - 2\alpha_1 - 1 = 0$. Given $\delta_1 - \alpha_2 = 4 \implies \delta_1 - \frac{1}{\delta_1} = 4 \implies \delta_1^2 - 4\delta_1 - 1 = 0$. Since $\alpha_1$ and $\delta_1$ are roots of the quartic equation,the quadratic factors are $(x^2 - 2x - 1)$ and $(x^2 - 4x - 1)$. Thus,$a x^4+b x^3+c x^2+d x+e = (x^2 - 2x - 1)(x^2 - 4x - 1)$. To find $a+b+c+d+e$,we substitute $x=1$: $a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = (1^2 - 2(1) - 1)(1^2 - 4(1) - 1)$. $a+b+c+d+e = (-2)(-4) = 8$.

Similar Questions

Let $p$ and $q$ be roots of the equation $x^2-2x+A=0$ and let $r$ and $s$ be the roots of the equation $x^2-18x+B=0$. If $p < q < r < s$ are in $A.P.$,then $A$ and $B$ are

If $\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$ for all real numbers $x$,then the range of the parameter $k$ is

Consider the equation $x^2 + \alpha x + \beta = 0$ having roots $\alpha, \beta$ such that $\alpha \neq \beta$. Also consider the inequality $||y - \beta| - \alpha| < \alpha$,then:

If $\tan \alpha$ and $\tan \beta$ are the roots of the equation $x^2 + px + q = 0$ $(p \neq 0)$,then:

The number of solutions of the equations $x+y+z=12$,$x^2+y^2+z^2=50$,and $x^3+y^3+z^3=216$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module