How do you find real zero of a function $f (x) = x^{2} - 8 x + 12$ $f(x) = x^2 - 8x + 12$ ?

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Answer 1 · 2015-09-11T01:50:42

Solve $f (x) = 0$ $f(x)=0$ we want only real (not imaginary) solutions.

A real zero of a function is a real number that, when substituted for $x$ $x$ makes $f (x) = 0$ $f(x)=0$

Solve $x^{2} - 8 x + 12 = 0$ $x^2-8x+12=0$

$x = 2$ $x=2$ , $x = 6$ $x=6$

The (real) zeros are 2 and 6.

How do you find real zero of a function f(x) = x^2 - 8x + 12f(x)=x2−8x+12f(x) = x^2 - 8x + 12?