If (2-i) is one of the roots of the equation | vedclass

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

Question

(C) Given the biquadratic equation $x^4-9x^3+31x^2-49x+30=0$. Since the coefficients are real,complex roots occur in conjugate pairs. Thus,if $(2-i)$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given the biquadratic equation $x^4-9x^3+31x^2-49x+30=0$. Since the coefficients are real,complex roots occur in conjugate pairs. Thus,if $(2-i)$ is a root,then $(2+i)$ is also a root. Let the four roots be $\alpha, \beta, (2-i),$ and $(2+i)$. The sum of the roots is $\alpha + \beta + (2-i) + (2+i) = -(-9)/1 = 9$. $\alpha + \beta + 4 = 9 \implies \alpha + \beta = 5$. The product of the roots is $\alpha \cdot \beta \cdot (2-i)(2+i) = 30/1 = 30$. Since $(2-i)(2+i) = 2^2 - i^2 = 4 + 1 = 5$,we have $\alpha \cdot \beta \cdot 5 = 30 \implies \alpha \cdot \beta = 6$. We have $\alpha + \beta = 5$ and $\alpha \cdot \beta = 6$. Solving these,we get $\alpha = 2$ and $\beta = 3$ (since $\alpha Therefore,$2\alpha - \beta = 2(2) - 3 = 4 - 3 = 1$.

If $(2-i)$ is one of the roots of the equation $x^4-9x^3+31x^2-49x+30=0$ and $\alpha, \beta$ $(\alpha < \beta)$ are its real roots,then $2\alpha-\beta=$

Similar Questions

If $a, b, c$ are distinct real numbers and $a^3 + b^3 + c^3 = 3abc$,then the equation $ax^2 + bx + c = 0$ has two roots,out of which one root is

Let $P(x)$ be a polynomial, which when divided by $(x-3)$ and $(x-5)$ leaves remainders $10$ and $6$, respectively. If the polynomial is divided by $(x-3)(x-5)$, then the remainder is

The sum of the fourth powers of the roots of the equation $16x^2-10x+1=0$ is

The sum of all the real values of $x$ satisfying the equation $2^{(x - 1)(x^2 + 5x - 50)} = 1$ is

Let $\alpha$ and $\beta$ be the roots of the equation $(x - a)(x - b) = c$,where $c \neq 0$. What are the roots of the equation $(x - \alpha)(x - \beta) + c = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module