Question

If alpha and beta are the zeros of polynomial 2x2-7x+5, then fine the value of alpha square divided by beta+beta square divided alpha

Answer 1

Given the roots of the equation:

$2x^2-7x+5=0$ are $\alpha ,\beta$

Compare it with:

$ax^2+bx+c=0$

So:

a=2

,b=-7

and c=5

We know that:

$\alpha+\beta=-b/a$

=7/2...............(1)

$\alpha*\beta=c/a$

=5/2.............(2)

Given

$\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}$

$\implies \frac{\alpha^3+\beta^3}{\alpha*\beta}$

$\implies \frac{(\alpha+\beta)(\alpha^2-\alpha*\beta+\beta^2)}{\alpha*\beta}$

$\implies \frac{(\alpha+\beta)([\alpha+\beta]^2-3\alpha*\beta)}{\alpha*\beta}$

From 1 and 2

$\implies \frac{(7/2)(49/4-3*5/2)}{5/2}$

$\implies \frac{(7/2)(19/4)}{5/2}$

$\implies \frac{133/8}{5/2}$

$\implies 133/8*2/5$

$\implies 133/20$

The answer is 133/20

Hope it helps

Answer 2

Answer:133/20

Step-by-step explanation:alpha and beta are the zeros of the polynomial

2x2-7x+5