Let the roots of the equation E_1 equiv x^3+x | vedclass

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

Question

(C) Let $x_1, x_2, x_3$ be the roots of the equation $E_1: x^3+x^2+lx+n=0$. From Vieta's formulas,$x_1+x_2+x_3 = -1$. Given $E_2: x^3+ax^2+bx+c=0$ is

Vedclass · Accepted Answer

$\sqrt{3}i, -\sqrt{3}i, \frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2}$

Vedclass · Answer

(C) Let $x_1, x_2, x_3$ be the roots of the equation $E_1: x^3+x^2+lx+n=0$. From Vieta's formulas,$x_1+x_2+x_3 = -1$. Given $E_2: x^3+ax^2+bx+c=0$ is a reciprocal equation of class one,so $c=1$ and $a=b$. Thus,$E_2: x^3+ax^2+ax+1=0$. The roots of $E_2$ are $\frac{x_i-1}{2}$. The sum of roots of $E_2$ is $\sum \frac{x_i-1}{2} = \frac{(x_1+x_2+x_3)-3}{2} = \frac{-1-3}{2} = -2$. From $E_2$,the sum of roots is $-a$,so $-a = -2 \Rightarrow a=2$. $E_2$ becomes $x^3+2x^2+2x+1 = (x+1)(x^2+x+1) = 0$. The roots of $E_2$ are $-1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$. Using $\frac{x_i-1}{2} = y_i$,we have $x_i = 2y_i+1$. For $y_1 = -1$,$x_1 = 2(-1)+1 = -1$. For $y_2 = \frac{-1+i\sqrt{3}}{2}$,$x_2 = 2(\frac{-1+i\sqrt{3}}{2})+1 = i\sqrt{3}$. For $y_3 = \frac{-1-i\sqrt{3}}{2}$,$x_3 = 2(\frac{-1-i\sqrt{3}}{2})+1 = -i\sqrt{3}$. The roots of $E_1$ are $\{-1, i\sqrt{3}, -i\sqrt{3}\}$. The roots of $E_2$ are $\{-1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}\}$. The common root is $-1$. Excluding the common root,the remaining roots are $\{i\sqrt{3}, -i\sqrt{3}, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}\}$.

Let the roots of the equation $E_1 \equiv x^3+x^2+lx+n=0$ be $x_i, (i=1, 2, 3)$ and the roots of $E_2 \equiv x^3+ax^2+bx+c=0$ be $\frac{x_i-1}{2}$. If the equation $E_2=0$ is a reciprocal equation of class one,then the roots of these two equations excluding the common roots are

Similar Questions

The number of all common roots of the equation $x^4-10x^3+37x^2-60x+36=0$ and the transformed equation obtained by increasing any two distinct roots of it by $1$,keeping the other two roots fixed,is

Let $a, b, c$ be real numbers with $a \ne 0$. If $\alpha$ is a root of $a^2x^2 + bx + c = 0$,$\beta$ is a root of $a^2x^2 - bx - c = 0$,and $0 < \alpha < \beta$,then the equation $a^2x^2 + 2bx + 2c = 0$ has a root $\gamma$ that always satisfies:

Statement-$I$: If $a + b + c = 0$ and $a, b, c$ are rational,then the roots of the equation $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ are rational.
Statement-$II$: The discriminant of $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ is a perfect square.

If the remainders of the polynomial $f(x)$ when divided by $(x + 1), (x - 2), (x + 2)$ are $6, 3, 15$ respectively,then the remainder of $f(x)$ when divided by $(x + 1)(x + 2)(x - 2)$ is:

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+2x^2-x-2=0$,then $\alpha^6+\beta^6+\gamma^6=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module