Question

if alpha and beta are the zeroes of the quadratic polynomial f(t)=t2-4t+3, find the value of alpha4*beta3+alpha3*beta4

Answer 1

Result :108

Hope you understand my answer and it nay helps you

Answer 2

"The value of $\left( { \alpha}^{ 4 }\quad \times \quad { \beta}^{ 3 } \right) \quad +\quad \left( { \alpha}^{ 3 }\quad \times \quad { \beta}^{ 4 } \right) \quad =\quad 108$

Solution:

$f\left( t \right) \quad =\quad { t }^{ 2 }\quad -\quad 4t\quad +\quad 3$

$=\quad { t }^{ 2 }\quad -\quad t\quad -\quad 3t\quad +3$

$=\quad t\left( t\quad -\quad 1 \right) \quad -3\left( t\quad -\quad 1 \right)$

$=\quad \left( t\quad -\quad 3 \right) \left( t\quad -\quad 1 \right)$

$\Rightarrow \quad t\quad =\quad 3;\quad 1$

So,

The zeros are 3 and 1

If alpha = 3

If beta = 1

From the question, we get

${ \alpha }^{ 4 }\quad \times \quad { \beta }^{ 3 }\quad \times \quad { \alpha }^{ 3 }\quad \times \quad { \beta }^{ 4 }\quad =\quad { 3 }^{ 4 }\quad \times \quad { 1 }^{ 3 }\quad \times \quad { 3 }^{ 3 }\quad \times \quad { 1 }^{ 4 }$

$\left( { \alpha }^{ 4 }\quad \times \quad { \beta }^{ 3 } \right) \quad +\quad \left( { \alpha }^{ 3 }\quad \times \quad { \beta }^{ 4 } \right) \quad =\quad \left( { 3 }^{ 4 }\quad \times \quad { 1 }^{ 3 } \right) \quad +\quad \left( { 3 }^{ 3 }\quad \times \quad { 1 }^{ 4 } \right)$

$=\quad 81\quad +\quad 27$

${ \alpha }^{ 4 }\quad \times \quad { \beta }^{ 3 }\quad \times \quad { \alpha }^{ 3 }\quad \times \quad { \beta }^{ 4 }\quad =\quad 108$"