if (2, 0) (0, 1) (4, 5) and (0, c) are con cyclic then find c
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Hi Dear....here is your answer.....
given,
Zeroes are 1/2+2√3 and 1/2-2√3
sum of zeroes = 1/2+2√3 + 1/2-2√3
= 1/2 + 1/2
= 1
product of zeroes = 1/2+2√3 x 1/2-2√3
= [ 1/2 ] ² - [ 2√3 ]²
= 1/4 - 12
= 1/4 - 48 / 4
= -47 /4
Quadratic equation
x² - [ sum of zeroes ]x + product of zeroes
=x² - x + -47/4............
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