Q. 1 E3.8( 13 Votes )

Find the zeroes o

Answer :

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

x2 – (√3 + 1)x + √3 = 0


To factorize the polynomial we have,


Sum of the value should be equal = -(√3 + 1)


Product should be equal to = 1 × √3


= √3


x2 – [2(√3 + 1)x]/2 + √3 = 0


x2 - 2(√3 + 1)x/2 + (√3 + 1)2/22 - (√3 + 1)2/22 + √3 = 0


[x - (√3 + 1)/2]2 - (3 + 1 + 2√3)/4 + √3 = 0


[x - (√3 + 1)/2]2 = (3 + 1 + 2√3 - 4√3)/4


[x - (√3 + 1)/2]2 = (√3 - 1)2/22


[x - (√3 + 1)/2] = ± (√3 - 1)/2


Solving for positive value,


x = (√3 - 1)/2 + (√3 + 1)/2


x = (√3 - 1 + √3 + 1)/2


x = 2√3/2 = √3


Solving for negative value,


x = -(√3 - 1)/2 + (√3 + 1)/2


x = (-√3 + 1 + √3 + 1)/2


x = 2/2 = 1


When we compare the above quadratic equation with the generalized one we get,


ax2 + bx + c = 0


a = 1, b = - (√3 + 1), c = √3


Sum of zeroes = -b / a


= (√3 + 1) / 1


= √3 + 1


Product of zeroes = c / a


= √3 / 1


= √3


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