Question

If alpha and beta are the zeros of quadratic polynomial P X equal to 3 x square - 6x plus 4 find the value of alpha and beta + beta upon alpha + 2 vision back at one upon alpha + 1 upon beta bracket close plus 3 alpha beta

Answer 1

Answer:

If alpha and beta are the zeros of quadratic polynomial P X equal to 3 x square - 6x plus 4 find the value of alpha and beta + beta upon alpha + 2 vision back at one upon alpha + 1 upon beta bracket close plus 3 alpha beta.

Answer 2

Answer:

$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac{1}{\alpha}+\frac{1}{\beta})+3\alpha\,\beta=8$

Step-by-step explanation:

Given polynomial, p(x) = 3x² - 6x + 4

α & β are zeroes of p(x)

using relationship of zero and coefficient of polynomial, we get

$\alpha+\beta=\frac{-(-6)}{3}=2$

$\alpha\,\beta=\frac{4}{3}$

$(\alpha+\beta)^2=\aplha^2+\alpha^2+2\alpha\,\beta$

$2^2=\aplha^2+\alpha^2+2\times\frac{4}{3}$

$\aplha^2+\alpha^2=4-\frac{8}{3}$

$\aplha^2+\alpha^2=\frac{4}{3}$

Now we are given with,

$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac{1}{\alpha}+\frac{1}{\beta})+3\alpha\,\beta$

$=\frac{\alpha^2+\alpha^2}{\alpha\,\beta}+2(\frac{\alpha+\beta}{\alpha\,\beta})+3\alpha\,\beta$

$=\frac{\frac{4}{3}}{\frac{4}{3}}+2(\frac{2}{\frac{4}{3}})+3\frac{4}{3}$

$=1+3+4$

$=8$

Therefore, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac{1}{\alpha}+\frac{1}{\beta})+3\alpha\,\beta=8$