If alpha and beta are two zeroes of the quadratic polynomial f(x) = 2x²-7x+3. Find the polynomial whose zeroes are 1/alpha and 1/beta.

Answer 1

After some manipulation and some carefully worked algebra with the quadratic formula for f(x) = 0, you come to find out (using either 1 plus or 1 minus, not always considering the plus/minus nature of square-roots)

α = ( 7 - √(43) ) / 2 ≈ 0.22, β = ( 7 + √(43) ) / 2 ≈ 6.78

1 / α = ( 7 + √(43) ) / 3 ≈ 4.52, 1 / β = ( 7 - √(43) ) / 3 ≈ 0.148

g(x) = ( x - 1 / α )( x - 1 / β )

........................................... = x² - ( ( 7 + √(43) ) / 3 + ( 7 - √(43) ) / 3 ) + ( ( 7 + √(43) ) / 3 )( ( 7 - √(43) ) / 3 )

........................................... = x² - 14/3 x + 6/9 = x² - 14/3 x + 2/3

Multiply everything by 3 (because scaling will not alter the roots/zeroes of the function) and you get:

g(x) = 3x² - 14x + 2

Where g(x) is the polynomial whose zeroes are 1 / α & 1 / β.

I hope this helps! Message me in the comments with any questions, comments, or concerns on what I did above!