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

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

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Answer 1 · 2016-03-05T18:04:33

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

Explanation:

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

$\Rightarrow 4 {(x + 3)}^{2} = 1$ $=> 4(x+3)^2 = 1$

$\Rightarrow {(x + 3)}^{2} = \frac{1}{4}$ $=> (x+3)^2 = 1/4$

$\Rightarrow x + 3 = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$ $=> x+3 = +-sqrt(1/4) = +-1/2$

$\Rightarrow x = - 3 \pm \frac{1}{2}$ $=>x = -3+-1/2$

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

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

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Explanation:

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

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