If alpha, beta are the roots of x^2 - 2x + 4 | vedclass

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

Question

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$. The roots are $\alpha, \beta$,so $\alpha + \beta = 2$ and $\alpha \beta = 4$. Since $\alpha$ is a

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$. The roots are $\alpha, \beta$,so $\alpha + \beta = 2$ and $\alpha \beta = 4$. Since $\alpha$ is a root,$\alpha^2 - 2\alpha + 4 = 0$,which implies $\alpha^2 = 2\alpha - 4$. Multiplying by $\alpha^n$,we get $\alpha^{n+2} = 2\alpha^{n+1} - 4\alpha^n$. Let $S_n = \alpha^n + \beta^n$. Then $S_n = 2S_{n-1} - 4S_{n-2}$. We know $S_0 = \alpha^0 + \beta^0 = 2$ and $S_1 = \alpha + \beta = 2$. $S_2 = 2S_1 - 4S_0 = 2(2) - 4(2) = 4 - 8 = -4$. $S_3 = 2S_2 - 4S_1 = 2(-4) - 4(2) = -8 - 8 = -16$. $S_4 = 2S_3 - 4S_2 = 2(-16) - 4(-4) = -32 + 16 = -16$. $S_5 = 2S_4 - 4S_3 = 2(-16) - 4(-16) = -32 + 64 = 32$. Thus,$\alpha^5 + \beta^5 = 32$.

If $\alpha, \beta$ are the roots of $x^2 - 2x + 4 = 0$,then $\alpha^5 + \beta^5$ is equal to

Similar Questions

Sachin and Rahul solve a quadratic equation. Sachin makes a mistake in writing the constant term and obtains roots $(4, 3)$. Rahul makes a mistake in writing the coefficient of $x$ and obtains roots $(3, 2)$. What are the correct roots of the equation?

Let $\alpha$ and $\beta$ be the roots of $x^{2}-6x-2=0$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$,then the value of $\frac{a_{10}-2a_{8}}{3a_{9}}$ is ..... .

If one root of the equation $x^2 + px + q = 0$ is the square of the other,then

For the equation $ax^2 + bx + c = 0$,the roots are $\alpha, \beta$,and for the equation $Ax^2 + Bx + C = 0$,the roots are $\alpha - k, \beta - k$. Then $\frac{B^2 - 4AC}{b^2 - 4ac} = \dots$

If $\alpha, \beta, \gamma$ are the roots of the equation $4x^3-3x^2+2x-1=0$,then $\alpha^3+\beta^3+\gamma^3=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module