Let alpha and beta be the roots of the equati | vedclass

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

Question

(C) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 6x - 2 = 0$.From the properties of roots,we have $\alpha + \beta =

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 6x - 2 = 0$.From the properties of roots,we have $\alpha + \beta = 6$ and $\alpha \beta = -2$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^2 - 2 = 6\alpha$$\beta^2 - 6\beta - 2 = 0 \implies \beta^2 - 2 = 6\beta$We are given $a_n = \alpha^n - \beta^n$. We need to evaluate $\frac{a_{10} - 2a_8}{2a_9}$.Substituting the expression for $a_n$:$\frac{a_{10} - 2a_8}{2a_9} = \frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$Using the relations $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$:$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$$= \frac{6\alpha^9 - 6\beta^9}{2(\alpha^9 - \beta^9)}$$= \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)} = \frac{6}{2} = 3$.

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - 6x - 2 = 0$. If $a_n = \alpha^n - \beta^n$ for $n \ge 1$,then the value of $\frac{a_{10} - 2a_8}{2a_9}$ is equal to:

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

Solve the given two equations and select the correct option.
$I.$ $\frac{12}{\sqrt{x}} - \frac{23}{\sqrt{x}} = 5\sqrt{x}$
$II.$ $\frac{\sqrt{y}}{12} - \frac{5\sqrt{y}}{12} = -\frac{1}{\sqrt{y}}$

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-x-1=0$. If $p_{k}=(\alpha)^{k}+(\beta)^{k}, k \geq 1$,then which one of the following statements is not true?

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+12x+36=0$
$II.$ $y^{2}+15y+56=0$

If the roots of the equation $5x^2 - 7x + k = 0$ are reciprocals of each other,then the value of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module