Question

if alpha and beta are the zeros of the polynomial
$3x {}^{2} - 6x + 4$
find the value of
$\alpha {}^{2} + \beta {}^{2}$
$\alpha {}^{2} - \beta {}^{2}$
$\alpha {}^{3} + \beta {}^{3}$


​

Answer 1

Step-by-step explanation:

Answer:

\begin{gathered}\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3\alpha\beta\\=8\end{gathered}

(

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

Answer 2

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