Question

Without finding the zeroes a and B of
polynomial p (x) = 6x2 - 13x + 6, find the
values of the following:(i) 1/alpha+1/beta (ii) alpha^2+beta^2 (iii) alpha^3+beta^3 (iv) alpha/beta+beta/alpha​

Answer 1

Explanation:

Given:

$p(x) = 6 {x}^{2} - 13x + 6$

To find:

$i) \frac{1}{ \alpha } + \frac{1}{ \beta } \\ \\ ii) { \alpha }^{2} + { \beta }^{2} \\ \\ iii) { \alpha }^{3} + { \beta }^{3} \\ \\ iv) \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$

Solution:

If $\alpha\:\:and\:\:\beta$ are zeros of quadratic polynomial $ax^2+bx+c$

then

$\alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a}$

According to the given quadratic polynomial

$\alpha + \beta = \frac{13}{6} ...eq1 \\ \\ \alpha \beta = \frac{6}{6} = 1...eq2$

i) To find $\frac{1}{ \alpha } + \frac{1}{ \beta}$

Divide eq2 by eq1

$\frac{ \alpha + \beta }{ \alpha \beta } = \frac{ \frac{13}{6} }{1} \\ \\ \frac{\cancel\alpha }{\cancel \alpha \beta } + \frac{\cancel \beta }{ \alpha \cancel\beta } = \frac{13}{6} \\ \\ \frac{1}{ \beta } + \frac{1}{ \alpha } = \frac{13}{6} \\ \\ or \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{13}{6}$

ii) To find ${ \alpha }^{2} + { \beta }^{2}$

Square eq1

$( { \alpha + \beta )}^{2} = \frac{169}{36} \\ \\ open \: whole \: square \\ \\ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{169}{36} \\ \\ put \: value \: of \: \alpha \beta \: from \: eq2 \\ \\ { \alpha }^{2} + { \beta }^{2} +2(1) = \frac{169}{36} \\ \\{ \alpha }^{2} + { \beta }^{2} = \frac{169}{36} -2\\\\ { \alpha }^{2} + { \beta }^{2} = \frac{169 - 72}{36} \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{97}{36}$

iii) To find ${ \alpha }^{3} + { \beta }^{3}$

Take cube of eq1

${( \alpha + \beta )}^{3} =\left ( \frac{13}{6} \right)^{3} \\ \\ open \: identity \: in \: left \: side \\ \\ { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta ) = \frac{2197}{216} \\ \\ put \: value \: of \: \alpha + \beta \: and \: \alpha \beta from \: eq1 \: and \: eq2 \\ \\ { \alpha }^{3} + { \beta }^{3} + 3(1) \left( \frac{13}{6} \right ) = \frac{2197}{216} \\ \\ { \alpha }^{3} + { \beta }^{3}= \frac{2197}{216} - \frac{39}{6} \\ \\ { \alpha }^{3} + { \beta }^{3}= \frac{2197 - 1404}{216} \\ \\ { \alpha }^{3} + { \beta }^{3}= \frac{793}{216}$

iv)To find $\frac{\alpha}{ \beta } + \frac{\beta}{ \alpha}$

Divide result of (ii) by eq2

$\frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } = \frac{97}{36} \\ \\ \frac{ { \alpha }^{ \cancel2} }{ \cancel\alpha \beta } + \frac{ { \beta }^{ \cancel2} }{ \alpha \cancel\beta } = \frac{97}{36} \\ \\ \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{97}{36}$

Final answer:

$i) \bold{\green{\frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{13}{6}} }$

$ii) \bold{\red{{ \alpha }^{2} + { \beta }^{2} = \frac{97}{36}} }$

$iii) \bold{\purple{{ \alpha }^{3} + { \beta }^{3}= \frac{793}{216} }}$

$iv) \bold{\pink{\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{97}{36} }}$

Hope it helps you.

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Answer 2

Explanation:

Solution:

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or

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