Question

alpha+beta=24, alpha-beta=8 find a polynomial whose zeroes are i) alpha^2 + beta^2 and alpha^2 - beta^2 ii) alpha,beta​

Answer 1

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

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:-

• Value of α = 16
• 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