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**integral** My instructor told us, to solve a quadratic inequality factor out it's roots to set up a test interval on a number line. Then, test to see at what positions the inequality is true.

My question:

$\displaystyle x^2+4x+4\geq 9

$

Descartes rule of signs:

0 positive roots

1 negative root

rational root theorem:

$\displaystyle

\textrm{roots}=\left [ -4,-2,-1 \right ]

$

Synthetic division:

............-4..0

-4......1..4..4..

........1..0..4---- $\displaystyle r=4\,\,\, \therefore $not a root

............-2..-4

-2.......1..4..4

.........1..2..0--------$\displaystyle r=4\,\,\,\therefore$ a root!

root=-2

$\displaystyle (x+2)(x+2)\geq 9$

But, this is a double root.

And because it is a double root my interval is... not applicable besides

But my textbook says the answer is:

$\displaystyle

\left ( -\infty, -5 \right ]\cup \left [ 1, \infty \right )$

Any help on this problem?

Also, if anyone has a more 'algebraic' way of solving a quadratic inequality, please post it. I hate guess and check problems.