if alpha and beta are the zeroes of 3x2-6x+4 . find the value of (alpha /beta+beta/alpha)+2(1/alpha+1/beta)+3alpha×beta
Answers
Answer:
Step-by-step explanation:
compare this with ax²+bx+c,we get
a = 3, b=-6, c=4,
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Answer:
(
β
α
+
α
β
)+2(
α
1
+
β
1
)+3αβ
=8
Step-by-step explanation:
\begin{gathered}Given\: \alpha \:and\:\beta\\are\: zeroes\:of\:3x^{2}-6x+4\end{gathered}
Givenαandβ
arezeroesof3x
2
−6x+4
compare this with ax²+bx+c,we get
a = 3, b=-6, c=4,
\begin{gathered}i) Sum\:of \:the \: zeroes\\=\frac{-b}{a}\\=\frac{-(-6)}{3}\\=\frac{6}{3}\\\alpha+\beta=2 --(1)\end{gathered}
i)Sumofthezeroes
=
a
−b
=
3
−(−6)
=
3
6
α+β=2−−(1)
\begin{gathered}ii) Product\:of\:the\: zeroes\\=\frac{c}{a}\\=\\alpha\beta=\frac{4}{3}---(2)\end{gathered}
ii)Productofthezeroes
=
a
c
=
alphaβ=
3
4
−−−(2)
\begin{gathered} iii)\alpha^{2}+\beta^{2}\\=\left(\alpha+\beta\right)^{2}-2\alpha\beta\\=\left(2\right)^{2}-2\times \frac{4}{3}\\=4-\frac{8}{3}\\=\frac{12-8}{3}\\=\frac{4}{3}--(3)\end{gathered}
iii)α
2
+β
2
=(α+β)
2
−2αβ
=(2)
2
−2×
3
4
=4−
3
8
=
3
12−8
=
3
4
−−(3)
\begin{gathered}Now,\\\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3\alpha\beta\\\end{gathered}
Now,
(
β
α
+
α
β
)+2(
α
1
+
β
1
)+3αβ
\begin{gathered}=\left(\frac{\alpha^{2}+\beta^{2}}{\alpha\beta}\right)+2\left(\frac{\beta+\alpha}{\alpha\beta}\right)+3\alpha\beta\\=\left(\frac{\frac{4}{3}}{\frac{4}{3}}\right)+2\left(\frac{2}{\frac{4}{3}}\right)+3\times \frac{4}{3}\end{gathered}
=(
αβ
α
2
+β
2
)+2(
αβ
β+α
)+3αβ
=(
3
4
3
4
)+2(
3
4
2
)+3×
3
4
= 1+2\times \frac{3}{2}+4=1+2×
2
3
+4
\begin{gathered}=1+3+4\\=8\end{gathered}
=1+3+4
=8