for the given polynomial p(x)=x^2-5x-1 if alpha and beeta are the zeros find the alpha^2+beeta+alpha×beeta^2
Answers
Step-by-step explanation:
α
2
+β
2
=
16
33
\alpha^2\beta+\beta ^2\alpha =\frac{-5}{16}α
2
β+β
2
α=
16
−5
Step-by-step explanation:
Given that alpha and beta are zeroes of polynomial p(x)=4x^2-5x-1p(x)=4x
2
−5x−1
we have to find the value of \alpha^2+\beta^2α
2
+β
2
and \alpha^2 \beta+\beta^2 \alphaα
2
β+β
2
α
Given polynomial is p(x)=4x^2-5x-1p(x)=4x
2
−5x−1
\begin{gathered}\alpha+\beta=\frac{-b}{a}=\frac{5}{4} \text{and}\\\\\alpha \beta =\frac{c}{a}=\frac{-1}{4}\end{gathered}
α+β=
a
−b
=
4
5
and
αβ=
a
c
=
4
−1
Now, \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=(\frac{5}{4})^2- 2(\frac{-1}{4})=\frac{25}{16}+\frac{1}{2}=\frac{33}{16}α
2
+β
2
=(α+β)
2
−2αβ=(
4
5
)
2
−2(
4
−1
)=
16
25
+
2
1
=
16
33
\alpha^2\beta+\beta ^2\alpha =\alpha\beta(\alpha+\beta)=\frac{-1}{4}(\frac{5}{4})=\frac{-5}{16}α
2
β+β
2
α=αβ(α+β)=
4
−1
(
4
5
)=
16
−5