Question

Find the value of alpha^2 + beta^2/alpha^-2 + beta^-2​

Answer 1

Answer:

HOW TO SOLVE?

$a {x}^{2} + bx + c = 0 \\ { \alpha }^{ - 2} + { \beta }^{ - 2} = \frac{1}{ { \alpha }^{2} } + \frac{1}{ { \beta }^{2} } \\ { \alpha }^{ - 2} + { \beta }^{ - 2} = \frac{ { \alpha }^{2} + { \beta }^{2} }{( { \alpha \beta) }^{2} } \\ to \: find \: value \: of \: = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha {}^{ - 2} + { \beta }^{ - 2} } \\ pu t \: value \: of \: { \alpha }^{ - 2} + { \beta }^{ - 2} \\ \frac{ \frac{ { \alpha }^{2} + { \beta }^{2} }{ { \alpha }^{2} + { \beta }^{2} } }{( \alpha \beta ) {}^{2} } = \frac{ \frac{1}{1} }{( { \alpha \beta )}^{2} } \\ = ( { \alpha \beta )}^{2} \\ put \: value \: of \: \alpha \beta = \frac{c}{a} \\( { \frac{c}{ {a}^{} } })^{2} = \frac{ {c}^{2} }{ {a}^{2} }$

It was a tricky question,

HAVE A NICE DAY!!!

HOPE IT HELPS YOU!!!

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

If (α + β)=24,( α - β) = 8 find a polynomial whose zeroes are i) α ² + β² and α² - β²

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

(α + β)=24,...(1)

( α - β) = 8 ...(2)

$\Large{\underline{\mathfrak{\bf{\pink{Find}}}}}$

A polynomial whose zeroes are (α ² + β²) and ( α² - β²)

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

Addition of equ(1) and equ(2)

➥ 2α = 32

➥ α = 32/2

➥ α = 16

Again, Keep value of " α " in equ(1)

➥ 16 + β = 24

➥ β = 24 - 16

➥ β = 8

Thus:-

1. Value of α = 16
2. Value of β = 8

Now,

(α ² + β²)

( keep value of α and β

➥ (16)²+(8)²

➥ 256+64

➥ 320

And,

( α² - β²)

( keep value of α and β )

➥ (16)²-(8)²

➥ 256 - 64

➥ 192

Now,

➩ Sum Of Zeroes = (α ² + β²)+( α² - β²)

➩ Sum Of Zeroes = (320)+(192)

➩ Sum Of Zeroes = 512

And,

• ➩ Product Of Zeroes = (α ² + β²)×( α² - β²)

• ➩ Product Of Zeroes = (320)×(192)

• ➩ Product Of Zeroes = 61,440

• Formula Quadratic Equation

[ x² -x(Sum of Zeroes)+(Product of zeroes ) = 0 ]

( Keep all values )

➩ [ x² - (512)x + (61,440) ] = 0

• This is required equation