Question

Find the zeros of the polynomial x² – 3 and verify the relationship between the zeros and the coefficients.

Answer 1

Dear Student:

Given: $x^{2} -3$

$a^{2} -b^{2} =(a-b)(a+b)$

$x^{2} -3=(x-\sqrt[2]{3} )(x+\sqrt[2]{3})=0$

$x-\sqrt[2]{3} =0$

x=√3

$x+\sqrt[2]{3} =0$

x=-√3

$ax^{2} +bx+c$

c= -3,a=1 and b=0

the sum of the zeros is zero which is -b/a=0

product of the zeros is -√3*√3=-3=c/a

Hope it helps

Thanks

With Regards

Answer 2

Let p(x) = x² - 3

We get zeroes of p(x) we will take

p(x) = 0

=> x² - 3 = 0

=> x² - ( √3 )² = 0

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We know the ,

Algebraic identity ,

a² - b² = ( a + b )( a - b )

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=> ( x + √3 )( x - √3 ) = 0

=> x + √3 = 0 or x - √3 = 0

The zeroes of given polynomial

are -√3 , √3 .

Compare x² - 3 with ax² + bx + c ,

we get

a = 1 , b = 0 , c = -3

Sum of the zeroes = -√3 + √3

= 0

= -b/a

Product of the zeroes = (√3)(-√3)

= -3

= c/a

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