The sum of the complex roots of the equation | vedclass

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

Question

(D) Given equation: $x^4-2x^3+x-380=0$. By checking the roots,for $x=5$: $5^4-2(5^3)+5-380 = 625-250+5-380 = 0$. So,$x=5$ is a root. For $x=-4$: $(-4)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given equation: $x^4-2x^3+x-380=0$. By checking the roots,for $x=5$: $5^4-2(5^3)+5-380 = 625-250+5-380 = 0$. So,$x=5$ is a root. For $x=-4$: $(-4)^4-2(-4)^3+(-4)-380 = 256+128-4-380 = 0$. So,$x=-4$ is a root. Let the four roots be $x_1, x_2, x_3, x_4$. We have $x_1=5$ and $x_2=-4$. Let $x_3$ and $x_4$ be the complex roots. From the relation between roots and coefficients,the sum of all roots is given by $-\frac{	ext{coefficient of } x^3}{	ext{coefficient of } x^4} = -\frac{-2}{1} = 2$. Thus,$x_1+x_2+x_3+x_4 = 2$. Substituting the known roots: $5-4+x_3+x_4 = 2$. $1+x_3+x_4 = 2$. Therefore,$x_3+x_4 = 1$.

The sum of the complex roots of the equation $x^4-2x^3+x-380=0$ is

Similar Questions

For the equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$,if $\alpha$ and $\beta$ are the roots,then $\alpha^2 + \beta^2 = \dots$

If the sum of two of the roots of $x^3+p x^2-q x+r=0$ is zero,then $pq$ is equal to

If the sum of any two roots of the equation $x^3+p x^2+q x+r=0$ is zero,then

Let $p(x)$ be a quadratic polynomial with constant term $1$. Suppose $p(x)$, when divided by $x-1$, leaves remainder $2$ and when divided by $x+1$, leaves remainder $4$. Then, the sum of the roots of $p(x)=0$ is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + x + 1 = 0$,then the value of $\alpha^3 \beta^3 \gamma^3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module