Question

Find the zeros of the polynomial and verify the relationship between the zeros and coefficients t square - 3

Answer 1

Let f(x)= t^2 - 3 .

For the zeros of the f(x).

t^2 -3=0

t^2 =3

t= +√3 or -√3 .

Thus the zeros of f(x) are +√3 and -√3 .

Now, sum of the zeros =+√3+(-√3)=0= -0/1=-(coefficient of t )/coefficient of t^2 .

Product of zeros = (-√3×√3)= -3 = -3/1 =( constant term )/coefficient of x^2 .

I hope it helps you dear .

Answer 2

Hello ...

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=> t^2 - 3 = 0
t^2 -(√3)^2 = 0
(t -√3 )(t +√3) = 0
t = +√3 and t =-√3

=> α+β = -b/a
√3 - √3 = 0 / 1
0 = 0

=> α×β = c/a
√3 × (-√3) = -3/ 1
-3 = -3

hence, verified

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