Question

If alpha and beta are zeroes of the polynomial f(t) = t^2 - 4t + 3. Find the value of:

(i) alpha^4 beta^3 + beta^4 alpha^3
(ii) alpha/beta + beta/alpha

Answer 1

$f(t) = {t}^{2} - 4t + 3 = 0 \\ \\ {t}^{2} - (3 + 1)t + 3 = 0 \\ \\ {t}^{2} - 3t - 1t + 3 = 0 \\ \\ t(t - 3) - 1(t - 3) = 0 \\ \\ (t - 1)(t - 3) = 0 \\ \\ t = 1 \: or \: t = 3 \\ \\ \alpha = 1 \: and \: \beta = 3$

$1. \: \alpha {}^{4} \times \beta {}^{4} + { \beta }^{4} + \alpha {}^{3} \\ \\ = {1}^{4} \times {3}^{4} + {3}^{4} \times 1 {}^{4} \\ \\ 1 \times 81 + 81 \times 1 \\ \\ 81 + 81 = 162$

$2. \: \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ \frac{1}{4} + \frac{4}{1} \\ \\ \frac{1 + 16}{4} = \frac{17}{4}$

Answer 2

$\mathfrak{Answer:-}$

$\mathsf{f(t) = {t}^{2} - 4t + 3}$

$\mathsf{f(t) = {t}^{2} - 4t + 3 = 0}$

$\mathsf{{t}^{2} - (3 + 1)t + 3 = 0}$

$\mathsf{{t}^{2} - 3t - 1t + 3 = 0}$

$\mathsf{t(t - 3) - 1(t - 3)= 0}$

$\mathsf{(t - 3) (t - 1)= 0}$

$\mathsf{t - 3 = 0 , t - 1 = 0}$

$\mathsf{t = 3, t = 1}$

$\mathfrak{ \alpha = 1 \: and \: \beta = 4}$

$\mathbf{1. \: { \alpha }^{4} \times { \beta }^{3} + { \beta }^{4} \times { \alpha }^{3} }$

$\mathbf{ = {1}^{4} \times {3}^{4} + {3}^{4} \times {1}^{4} }$

$\mathbf{ = 1 \times 81 + 81\times 1}$

$\mathbf{ = 81 + 81}$

$\mathbf{ = 162}$

$\mathbf{ 2. \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }}$

$\mathbf{ = \frac{1}{4} + \frac{4}{1}}$

$\mathbf{ = \frac{1 + 16}{4}}$

$\mathbf{ = \frac{17}{4}}$