How do you find all the zeros of $f(x) = 4(x + 3)^2 - 1$ ?

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Answer 1 · 2016-03-05T18:04:33

Set $f(x) = 0$ and solve for $x$ to find that $f(x)$ has zeroes at $x=-7/2$ and $x=-5/2$

$4(x+3)^2-1 = 0$

$=> 4(x+3)^2 = 1$

$=> (x+3)^2 = 1/4$

$=> x+3 = +-sqrt(1/4) = +-1/2$

$=>x = -3+-1/2$

$:. x = -7/2$ or $x = -5/2$

How do you find all the zeros of f(x) = 4(x + 3)^2 - 1f(x) = 4(x + 3)^2 - 1?