If alpha, beta, gamma are roots of x^3 - 2x^2 | vedclass

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

Question

(A) Given the equation $x^3 - 2x^2 + 3x - 2 = 0$. By inspection,$x = 1$ is a root. Dividing by $(x - 1)$,we get $(x - 1)(x^2 - x + 2) = 0$. Thus,$\alp

Vedclass · Accepted Answer

$\frac{13}{4}$

Vedclass · Answer

(A) Given the equation $x^3 - 2x^2 + 3x - 2 = 0$. By inspection,$x = 1$ is a root. Dividing by $(x - 1)$,we get $(x - 1)(x^2 - x + 2) = 0$. Thus,$\alpha = 1$ and $\beta, \gamma$ are roots of $x^2 - x + 2 = 0$. From the quadratic equation,$\beta + \gamma = 1$ and $\beta \gamma = 2$. We need to evaluate $S = \frac{\alpha \beta}{\alpha + \beta} + \frac{\alpha \gamma}{\alpha + \gamma} + \frac{\beta \gamma}{\beta + \gamma}$. Substituting $\alpha = 1$: $S = \frac{\beta}{1 + \beta} + \frac{\gamma}{1 + \gamma} + \frac{2}{1} = \frac{\beta(1 + \gamma) + \gamma(1 + \beta)}{(1 + \beta)(1 + \gamma)} + 2$. $S = \frac{\beta + \beta \gamma + \gamma + \beta \gamma}{1 + (\beta + \gamma) + \beta \gamma} + 2 = \frac{(\beta + \gamma) + 2(\beta \gamma)}{1 + (\beta + \gamma) + \beta \gamma} + 2$. Substituting values: $S = \frac{1 + 2(2)}{1 + 1 + 2} + 2 = \frac{5}{4} + 2 = \frac{13}{4}$.

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

Similar Questions

Let $a, b, c$ be the roots of the equation $x^3 + 8x + 1 = 0$. Then the value of $\frac{bc}{(8b + 1)(8c + 1)} + \frac{ac}{(8a + 1)(8c + 1)} + \frac{ab}{(8a + 1)(8b + 1)}$ is equal to

The harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is:

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

If $\alpha, \beta$ are the roots of $x^2+ax+2=0$ and $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots of $x^2-bx+c=0$,then $\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right) = $

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+a x^2-b x+c=0$,then $\sum \beta^2(\gamma+\alpha) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module