Question

pls solve it fast....
Find the zeros of x^2'- 7​

Answer 1

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p(x)= x²-7

⟹(x+√7)(x-√7)= 0

Therefore,

⟹x+√7= 0 and x-√7= 0

⟹x= -√7 and x= √7

Therefore, -√7 and √7 are the required zeros

Answer 2

$x^{2} - 7 = (x+\sqrt{7} )(x-\sqrt{7} )\\Since, a^{2} - b^{2}  = (a+b)(a-b)\\\\x+\sqrt{7} = 0\\=> x = -\sqrt{7} \\x-\sqrt{7} = 0\\=>x = \sqrt{7}$

Therefore, the two zeroes are $\sqrt{7}$ and $-\sqrt{7}$