Question

Without finding the zeroes a and B of
polynomial p (x) = x^2-5x+6,find the value of (i)alpha/beta+beta/alpha (ii) alpha^2+beta^2 (iii) alpha^3+beta^3 (iv) alpha^4+beta^4​

Answer 1

Step-by-step-Explanation:

Given:$p(x) = {x}^{2} - 5x + 6$

To find:

$i) \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ ii) { \alpha }^{2} + { \beta }^{2} \\ \\ iii) { \alpha }^{3} + { \beta }^{3} \\ \\ iv) { \alpha }^{4} + { \beta }^{4}$

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$\alpha + \beta = 5...eq1 \\ \\ \alpha \beta = 6...eq2$

ii) To find ${ \alpha }^{2} + { \beta }^{2}$Square eq1

$( { \alpha + \beta )}^{2} =25 \\ \\ open \: whole \: square \\ \\ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta =25 \\ \\ put \: value \: of \: \alpha \beta \: from \: eq2 \\ \\ { \alpha }^{2} + { \beta }^{2} +2(6) = 25 \\ \\{ \alpha }^{2} + { \beta }^{2} = 25 -12\\\\ { \alpha }^{2} + { \beta }^{2} = 13$

iii) To find ${ \alpha }^{3} + { \beta }^{3}$Take cube of eq1${( \alpha + \beta )}^{3} =( 5)^{3} \\ \\ open \: identity \: in \: left \: side \\ \\ { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta ) = 125 \\ \\ put \: value \: of \: \alpha + \beta \: and \: \alpha \beta from \: eq1 \: and \: eq2 \\ \\ { \alpha }^{3} + { \beta }^{3} + 3(6)(5) = 125 \\ \\ { \alpha }^{3} + { \beta }^{3}= 125 - 90 \\ \\ { \alpha }^{3} + { \beta }^{3}= 35$

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

Divide result of (ii) by eq2

$\frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } = \frac{13}{6} \\ \\ \frac{ { \alpha }^{ \cancel2} }{ \cancel\alpha \beta } + \frac{ { \beta }^{ \cancel2} }{ \alpha \cancel\beta } = \frac{13}{6} \\ \\ \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{13}{6}$

iv) To find ${ \alpha }^{4} + { \beta }^{4}$

Take fourth power of eq1

${( \alpha + \beta )}^{4} = {(5)}^{4} \\ \\ open \: identity \: in \: LHS \\ \\ { \alpha }^{4} + 4 { \alpha }^{3} \beta + 6 { \alpha }^{2} { \beta }^{2} + 4 \alpha { \beta }^{3} + { \beta }^{4} = 625 \\ \\ { \alpha }^{4} + { \beta }^{4} + 6 { \alpha }^{2} { \beta }^{2} + 4 \alpha \beta ( { \alpha }^{2} + { \beta }^{2} ) = 625$

put the value of $\alpha\beta$ from eq2 and result of part(ii)

${ \alpha }^{4} + { \beta }^{4} + 6 ({ \alpha \beta })^{2} + 4 \alpha \beta ( { \alpha }^{2} + { \beta }^{2} ) = 625 \\ \\ { \alpha }^{4} + { \beta }^{4} + 6 (6)^{2} + 4 (6) (13 ) = 625 \\ \\ { \alpha }^{4} + { \beta }^{4} + 216 + 312 = 625 \\ \\ { \alpha }^{4} + { \beta }^{4} + 528 = 625 \\ \\ { \alpha }^{4} + { \beta }^{4} = 625 - 528 \\ \\ { \alpha }^{4} + { \beta }^{4} = 97$

Final answer:

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

$ii) \bold{\red{{ \alpha }^{2} + { \beta }^{2} = 13} }$

$iii) \bold{\purple{{ \alpha }^{3} + { \beta }^{3}= 35 }}$

$iv) \bold{\pink{{ \alpha }^{4} + { \beta }^{4} = 97 }}$

Hope it helps you.

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