alpha and beta are zeros of x^2-2x+3 find a polynomial the zeros are alpha-1/alpha+1 and beta-1/beta+1
Answers
Answer:
3x² - 2x + 1
Step-by-step explanation:
By Vieta's formulas:
α + β = 2 and αβ = 3
For our answer, we look for a quadratic:
x² - Ax + B
where
A = ( α - 1 ) / ( α + 1 ) + ( β - 1 ) / ( β + 1 )
B = ( α - 1 ) ( β - 1 ) / ( α + 1 ) ( β + 1 )
The value of A
A = ( α - 1 ) / ( α + 1 ) + ( β - 1 ) / ( β + 1 )
= [ ( α - 1 ) ( β + 1 ) + ( α + 1 ) ( β - 1 ) ] / ( α + 1 ) ( β + 1 )
= ( αβ + α - β - 1 + αβ - α + β - 1 ) / ( αβ + α + β + 1 )
= ( 2αβ - 2 ) / ( αβ + α + β + 1 )
= ( 2×3 - 2 ) / ( 3 + 2 + 1 )
= 4 / 6
= 2 / 3
The value of B
B = ( α - 1 ) ( β - 1 ) / ( α + 1 ) ( β + 1 )
= ( αβ - ( α + β ) + 1 ) / ( αβ + α + β + 1 )
= ( 3 - 2 + 1 ) / ( 3 + 2 + 1 )
= 2 / 6
= 1 / 3
Our answer
So a quadratic polynomial that answers the question is:
x² - (2/3)x + 1/3
However, we can choose a nicer looking answer by noting that multiplying by a constant doesn't change the zeros. So we multiply this by 3 to give a nicer answer:
3x² - 2x + 1