if alpha and beta are the zeros of the quadratic polynomial f(t)=t2-4t+3 find the value of alpha4beta3+alpha3beta4
Answers
Step-by-step explanation:
)=108
Solution:
f\left( t \right) \quad =\quad { t }^{ 2 }\quad -\quad 4t\quad +\quad 3f(t)=t
2
−4t+3
=\quad { t }^{ 2 }\quad -\quad t\quad -\quad 3t\quad +3=t
2
−t−3t+3
=\quad t\left( t\quad -\quad 1 \right) \quad -3\left( t\quad -\quad 1 \right)=t(t−1)−3(t−1)
=\quad \left( t\quad -\quad 3 \right) \left( t\quad -\quad 1 \right)=(t−3)(t−1)
\Rightarrow \quad t\quad =\quad 3;\quad 1⇒t=3;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 }α
4
×β
3
×α
3
×β
4
=3
4
×1
3
×3
3
×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)(α
4
×β
3
)+(α
3
×β
4
)=(3
4
×1
3
)+(3
3
×1
4
)
=\quad 81\quad +\quad 27=81+27
{ \alpha }^{ 4 }\quad \times \quad { \beta }^{ 3 }\quad \times \quad { \alpha }^{ 3 }\quad \times \quad { \beta }^{ 4 }\quad =\quad 108α
4
×β
3
×α
3
×β
4
=108 "
Answer:
22 ok am i i right plzz tell
Step-by-step explanation:
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