power 4 + x power cube + b x power square + 3 x + the the sigma of alpha square plus beta square plus gamma square is equals to how much
Answers
Answer:
4+x³+bx²+3x
Step-by-step explanation:
\alpha ^2+\beta ^2+\gamma ^2+\delta ^2=a^2-2bα
2
+β
2
+γ
2
+δ
2
=a
2
−2b
Step-by-step explanation:
Given that
x^4+ax^3+bx^2+cx+d=0x
4
+ax
3
+bx
2
+cx+d=0
This is the 4th order equation .
We have to find
\sum \alpha ^2=\alpha ^2+\beta ^2+\gamma ^2+\delta ^2∑α
2
=α
2
+β
2
+γ
2
+δ
2
As we know that
\left ( a+b+c +d\right )^2=a^2+b^2+c^2+2\left ( ab+bc+ac+ad+db+cd \right )(a+b+c+d)
2
=a
2
+b
2
+c
2
+2(ab+bc+ac+ad+db+cd)
α+β+γ+δ= -a
αβ+βγ+γδ+δα+βδ+αγ= b
So
\alpha ^2+\beta ^2+\gamma ^2+\delta ^2=\left ( \alpha +\beta +\gamma +\delta \right )^2-2\left ( \alpha \beta +\beta \gamma +\gamma \delta +\delta \alpha+\alpha \gamma +\beta \delta \right )α
2
+β
2
+γ
2
+δ
2
=(α+β+γ+δ)
2
−2(αβ+βγ+γδ+δα+αγ+βδ)
Now by putting the values
\alpha ^2+\beta ^2+\gamma ^2+\delta ^2=a^2-2bα
2
+β
2
+γ
2
+δ
2
=a
2
−2b
So
\alpha ^2+\beta ^2+\gamma ^2+\delta ^2=a^2-2bα
2
+β
2
+γ
2
+δ
2
=a
2
−2b