If the sum of two roots of the equation x^4+2 | vedclass

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

Question

(B) Let the roots of the equation $x^4+2x^3-7x^2-8x+12=0$ be $\alpha, \beta, \gamma, \delta$. Given that the sum of two roots is zero,let $\alpha + \b

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Let the roots of the equation $x^4+2x^3-7x^2-8x+12=0$ be $\alpha, \beta, \gamma, \delta$. Given that the sum of two roots is zero,let $\alpha + \beta = 0$,which implies $\beta = -\alpha$. From Vieta's formulas,the sum of roots is $\alpha + \beta + \gamma + \delta = -2$. Since $\alpha + \beta = 0$,we have $\gamma + \delta = -2$. The product of roots is $\alpha \beta \gamma \delta = 12$. Since $\beta = -\alpha$,we have $-\alpha^2 \gamma \delta = 12$,or $\alpha^2 \gamma \delta = -12$. Also,the sum of roots taken two at a time is $\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = -7$. Substituting $\beta = -\alpha$,we get $-\alpha^2 + \alpha(\gamma + \delta) - \alpha(\gamma + \delta) + \gamma \delta = -7$. This simplifies to $-\alpha^2 + \gamma \delta = -7$,so $\gamma \delta = \alpha^2 - 7$. Substituting $\gamma \delta$ into the product equation: $\alpha^2(\alpha^2 - 7) = -12$,which gives $\alpha^4 - 7\alpha^2 + 12 = 0$. Let $y = \alpha^2$,then $y^2 - 7y + 12 = 0$,so $(y-3)(y-4) = 0$. Thus $\alpha^2 = 3$ or $\alpha^2 = 4$. If $\alpha^2 = 4$,then $\gamma \delta = 4 - 7 = -3$. We know $\gamma + \delta = -2$. The sum of squares $\gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma \delta = (-2)^2 - 2(-3) = 4 + 6 = 10$. If $\alpha^2 = 3$,then $\gamma \delta = 3 - 7 = -4$. Then $\gamma^2 + \delta^2 = (-2)^2 - 2(-4) = 4 + 8 = 12$. Checking the original equation,the roots are $2, -2, 1, -3$. The sum of two roots being zero is $2 + (-2) = 0$. The other two roots are $1$ and $-3$. The sum of their squares is $1^2 + (-3)^2 = 1 + 9 = 10$.

If the sum of two roots of the equation $x^4+2x^3-7x^2-8x+12=0$ is zero,then the sum of the squares of the other two roots is

Similar Questions

If the sum of the roots of the equation $x^2 + px + q = 0$ is equal to the sum of their squares,then

Suppose $\alpha, \beta, \gamma$ are the roots of $x^3+x^2+x+2=0$. Then,the value of $\left(\frac{\alpha+\beta-2 \gamma}{\gamma}\right)\left(\frac{\beta+\gamma-2 \alpha}{\alpha}\right)\left(\frac{\gamma+\alpha-2 \beta}{\beta}\right)$ is

Let $\alpha, \beta$ be the roots of $x^{2}-x-1=0$ and $S_{n}=\alpha^{n}+\beta^{n}$ for all integers $n \geq 1$. Then, for every integer $n \geq 2$, which of the following is true?

If ${x^2} + px + q = 0$ is the quadratic equation whose roots are $a - 2$ and $b - 2$,where $a$ and $b$ are the roots of ${x^2} - 3x + 1 = 0$,then:

Let $\alpha$ and $\beta$ be two real numbers such that $\alpha+\beta=1$ and $\alpha \beta=-1$. Let $p_{n}=\alpha^{n}+\beta^{n}$,$p_{n-1}=11$ and $p_{n+1}=29$ for some integer $n \geq 1$. Then,the value of $p_{n}^{2}$ is .... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module