Answers

The Brainliest Answer!
2016-05-05T11:49:31+05:30
 x^{2} -1=0 \\  \\ (x+1)(x-1)=0   [a ^{2}- b^{2} =(a+b)(a-b)] \\  \\ Either: \\ (x-1)=0 \\ or \\ (x+1)=0 \\ \\ So: \\ x=1 \\ or \\ x=(-1) \\ 

So, the zeros are 1, (-1).

ALITER:
By the Quadratic formula:
 \alpha = \frac{-b+ \sqrt{ b^{2} -4ac} }{2a}  \\  \\  \beta \frac{-b- \sqrt{ b^{2} -4ac} }{2a}
Here: 
a = coefiicient of  x^{2}       (Here : 1)
b = coefficient of x                              (Here : 0)
c = constant                                       (Here : -1)
Solving:
 \alpha = \frac{-0+ \sqrt{ 0^{2} -4(1)(-1)} }{2(1)}=  \frac{ \sqrt{4} }{2} = 1 \\  \\  \beta =   \frac{-0- \sqrt{ 0^{2} -4(1)(-1)} }{2(1)}= \frac{- \sqrt{4} }{2} = -1
1 5 1
but actually the number of zeroz are 0
no real number of zeros
@helpinghand
Read it as (x^2 + 0.x - 1) and solve it by the Quadratic formula. You will get the same answer.
I will add that.....just a minute