If a nand b are the zeros of the polynomial p(x) = 3x2 - 14x + 15,find the value of A2 + b2
Answers
Answer:
The value of \alpha^2+\beta^2=\frac{106}{9}α
2
+β
2
=
9
106
Step-by-step explanation:
p(x) = 3x^2 - 14x +15p(x)=3x
2
−14x+15
\alphaα and \betaβ are zeroes
General equation : ax^2+bx+c=0ax
2
+bx+c=0
Sum of zeroes = \frac{-b}{a}
a
−b
Product of zeroes = \frac{c}{a}
a
c
Sum of zeroes of given equation : \alpha+\beta=\frac{14}{3}α+β=
3
14
Product of zeroes of given equation : \alpha\beta =\frac{15}{3}=5αβ=
3
15
=5
we are supposed to find \alpha^2+\beta^2α
2
+β
2
Formula : \begin{gathered}(a+b)^2=a^2+b^2+2ab\\(a+b)^2-2ab=a^2+b^2\end{gathered}
(a+b)
2
=a
2
+b
2
+2ab
(a+b)
2
−2ab=a
2
+b
2
(\alpha+\beta)^2-2 \alpha \beta=\alpha^2+\beta^2(α+β)
2
−2αβ=α
2
+β
2
Substitute the values:
\begin{gathered}(\frac{14}{3})^2-2(5)=\alpha^2+\beta^2\\\\\frac{106}{9}=\alpha^2+\beta^2\end{gathered}
(
3
14
)
2
−2(5)=α
2
+β
2
9
106
=α
2
+β
2
Hence the value of \alpha^2+\beta^2=\frac{106}{9}α
2
+β
2
=
9
106