find the zeros of f(x)=x^2+2

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f(x) = x² + 2 y = x² + 2 (rewrite f(x) as y) x² + 2 = 0 (set quadratic equation to zero) Standard form of quadratic equation: ax² + bx + c = 0 Use quadratic formula to solve: x = (-b +/- (b² - 4ac)^1/2 ) / 2a Coefficients (a,b,c) for x² + 2 = 0: a = 1; b = 0; c = 2 Plug in coefficients (a,b,c) into quadratic formula and solve: x = (-0 +/- (0² - 4(1)(2))^1/2) / 2(1) x = +/- (-8)^1/2 / 2 x = +/- (4)^1/2(2)^1/2(-1)^1/2 / 2 x = +/- 2(2i) / 2 There are 2 imaginary roots (zeroes) of our function, f(x) = x² + 2:  +2i and -2i