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Given: $f(x)=x^{2}+8 x+16$

  1. Solve for $x$ if $f(x)>0$
  2. For which values of $p$ will $f(x)=p$ have TWO unequal negative roots?
in Mathematics by Diamond (74,866 points) | 5 views

2 Answers

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Best answer
\begin{array}{r}
x^{2}+8 x+16>0 \\
(x+4)(x+4)>0
\end{array}

The function values remain positive $x \in R, x \neq-4$
by Diamond (74,866 points)
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\begin{array}{l}
x^{2}+8 x+16>0 \\
(x+4)(x+4)>0 \\
x \in R, x \neq-4 \quad \text { or } \\
x \in(-\infty ;-4) \text { or } x \in(-4 ; \infty) \text { or } \\
x<-4 \text { or } x>-4
\end{array}
by Diamond (74,866 points)

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