Question

if alpha and beta are the zeroes of the polynomial p(x)=3xsquare-5x+6, find (i) alpha/beta+beta/alpha (ii) alpha cube +beta cube

Answer 1

Hey There!


Here we are given the polynomial:

$p(x)=3x^2-5x+6$

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


So, 
Sum of zeros is:

$\alpha + \beta = -\frac{(-5)}{3} \\ \\ \\ \implies \boxed{\alpha + \beta = \frac{5}{3}}$

Also, Product of zeros is:

$\alpha \beta = \frac{6}{3} \\ \\ \\ \implies \boxed{\alpha \beta = 2}$


Now, with this info, we can find whatever we need:



$1) \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \\ \\ \\ = \frac{\alpha^2 + \beta^2}{\alpha \beta} \\ \\ \\ = \frac{\alpha^2 + \beta^2 + 2 \alpha \beta - 2 \alpha \beta}{\alpha \beta} \\ \\ \\ = \frac{(\alpha + \beta )^2 -2 \alpha \beta}{\alpha \beta} \\ \\ \\ = \frac{\left(\frac{5}{3}\right)^2-2(2)}{2} \\ \\ \\ = \frac{\frac{25}{9}-4}{2} \\ \\ \\ = \frac{25-36}{18} \\ \\ \\ = -\frac{11}{18} \\ \\ \\ \\ \implies \boxed{\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -\frac{11}{18}}$


$2) \alpha^3 + \beta^3 \\ \\ \\ = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2) \\ \\ \\ = (\alpha + \beta)(\alpha^2 + 2 \alpha \beta + \beta^2 - 3\alpha \beta) \\ \\ \\ = (\alpha + \beta)((\alpha + \beta)^2 - 3 \alpha \beta) \\ \\ \\ = \frac{5}{3} \times \left( \left( \frac{5}{3} \right)^2 - 3(2) \right) \\ \\ \\ = \frac{5}{3} \left( \frac{25}{9} - 6 \right) \\ \\ \\ = \frac{5}{3} \times \frac{25-54}{9} \\ \\ \\ \implies = -\frac{29}{9} \times \frac{5}{3} \\ \\ \\ = -\frac{145}{27} \\ \\ \\ \\ \implies \boxed{\alpha^3 + \beta^3 = -\frac{145}{27}}$


Hope it helps
Purva
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