# How do you solve x^2+3x-18>=0 by algebraically?

Jan 8, 2017

$\left\{x : x \in \left[3 \text{ to } + \infty\right)\right\}$

$\left\{x : x \in \left[- 6 \text{ to } - \infty\right)\right\}$

#### Explanation:

Solve for ${x}^{2} + 3 x - 18 = 0$ and the resulting values for $x$ form the reference points for 2 sets of values (domains -> input)

Set ${x}^{2} + 3 x - 18 = 0$

Note that $3 \times 6 = 18 \text{ and that } 6 - 3 = 3$ giving:

$\left(x + 6\right) \left(x - 3\right) = {x}^{2} + 3 x - 18 = 0$

So for this condition $x = - 6 \text{ and } x = + 3$
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As the ${x}^{2}$ term is positive the graph is of form $\cup$. So we are looking for all those values of $x$ that relate to the plot where it is cuts and is above the x-axis. In other words they will be 'traveling' away from the origin in both the positive and negative direction.

$x \le - 6 \text{ and } x \ge 3$
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The set $x$ such that it is all values from and including -6 to approaching $- \infty$

The set $x$ such that it is all values from and including +3 to approaching $+ \infty$

$\left\{x : x \in \left[3 \text{ to } + \infty\right)\right\}$

$\left\{x : x \in \left[- 6 \text{ to } - \infty\right)\right\}$