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arrow_back Prove that, for all values of \(x\), \[ x^{2}+4 x+12>x+3 \]

by Platinum
(106,844 points)
in Mathematics
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Prove that, for all values of \(x\),
\[
x^{2}+4 x+12>x+3
\]

1 Answer

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Best answer
\[
\begin{aligned}
&x^{2}+4 x+12>x+3 \\
&x^{2}+3 x+9>0 \\
&\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}+9>0 \\
&\left(x+\frac{3}{2}\right)^{2}+\frac{27}{4}>0
\end{aligned}
\]
Min point at \(\left(-\frac{3}{2}, \frac{27}{4}\right)\)
\(\therefore x^{2}+4 x+12>x+3\) for all values of \(x\)
by Platinum
(106,844 points)

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