Question

if alpha and beta are the zeros of the polynomial 6x²+x-2.find the value of (alpha/beta + beta/alpha)

Answer 1

Here we are given the polynomial:

$6x^2+x-2$

Also, $\alpha$ and $\beta$ are zeros. 


Sum of zeros is:
$\alpha + \beta = -\frac{\text{Coefficient of x}}{\text{Coefficient of }x^2} \\ \\ \\ \implies \boxed{\alpha + \beta = -\frac{1}{6}}$

And, Product of zeros is:

$\alpha \beta = \frac{\text{Constant Term}}{\text{Coefficient of }x^2} \\ \\ \\ \implies \alpha \beta = \frac{-2}{6} \\ \\ \\ \implies \boxed{\alpha \beta = -\frac{1}{3}}$


Now, we can find our answer as follows:

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} \\ \\ \\ = \frac{\alpha^2 + \beta^2}{\alpha \beta} \\ \\ \\ = \frac{\alpha^2 + 2\alpha \beta + \beta^2 - 2\alpha \beta}{\alpha \beta} \\ \\ \\ = \frac{(\alpha + \beta)^2-2\alpha \beta}{\alpha \beta} \\ \\ \\ = \frac{(\alpha + \beta)^2}{\alpha \beta} - 2 \\ \\ \\ = \frac{\left( -\frac{1}{6}\right)^2}{-\frac{1}{3}} - 2 \\ \\ \\ = \left(-\frac{1}{36} \times 3\right)-2 \\ \\ \\ = -\frac{1}{12} - 2 \\ \\ \\ = \frac{-1-24}{12} \\ \\ \\ = -\frac{25}{12} \\ \\ \\ \\ \\ \implies \boxed{\frac{\alpha}{\beta}+\frac{\beta}{\alpha} = -\frac{25}{12}}$

Hope it helps
Purva
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