Question

p(x)=x-x²-1 ,Find the number of zeros of the given polynomial.

Answer 1

hope it will helps U............

Answer 2

Solution :

Given Quadratic polynomial

p(x) = x - x² - 1

= -x² + x - 1

Let p(x) = 0

=> -x² + x - 1 = 0

=> x² - x + 1 = 0

=> x² - x = -1

=> x² - 2*x*(1/2) = -1

=> x² - 2*x*(1/2) + (1/2)² = -1 + (1/2)²

=> ( x - 1/2 )² = -1 + 1/4

=> ( x - 1/2 )² = ( -4 + 1 )/4

=> ( x - 1/2 )² = -3/4

=> x - 1/2 = ± √(-3)/2

=> x = 1/2 ± √(-3)/2

=> x = [ 1 ± √(-3) ]/2

Therefore ,

[ 1±√(-3)]/2 are two zeroes of p(x)

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