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More answers

y = x2 - 2

Set y = 0

0 = x2 - 2

0= x2 - (sq. root of 2)2

0 = [x - (sq. root of 2)][x + (sq. root of 2)]

x - (sq. root of 2) = 0 or x + (sq. root of 2) = 0

x = (sq. root of 2) or x = -(sq. root of 2)

Thus, the x-intercepts are +&- sq. root of 2.

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Q: The x intercepts of y equals x2-2?
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