Question

If alpha and beta are the zero of the polynomial t^2 - 5t +3 then find alpha^3peta^4+alpha^4beta^3

Answer 1

Hey!


Given :- t² - 5t + 3

• Sum of Zeros =
$\frac{ - coefficient \: of \: x}{coefficient \: of \: \: {x}^{2} }$


$\alpha + \beta = \frac{5}{1}$


• Product of Zeros =
$\frac{ constant \: \: term}{coeffic. \: \: \: of \: {x}^{2} }$


$\alpha \beta = \frac{3}{1}$


# To find :-

${ \alpha }^{3} { \beta }^{4} + { \alpha }^{4} { \beta }^{3}$


${ \alpha }^{3} { \beta }^{3} ( \beta + \alpha )$


${( \alpha \beta )}^{3} ( \alpha + \beta )$


$( {3)}^{3} \times 5$

= 135


Hence , value is 135 !!