# Determine the interval whereby 6x^2 + 44x + 70 ≥ 0?

Oct 14, 2016

$x \in \left(- \infty , - 5\right] \cup \left[- \frac{7}{3} , \infty\right)$

#### Explanation:

$6 {x}^{2} + 44 x + 70 = \left(2 x + 10\right) \left(3 x + 7\right) \ge 0$

The product of two values is positive only if both values are positive or both values are negative. Thus

$2 x + 10 \ge 0 \mathmr{and} 3 x + 7 \ge 0$
or
$2 x + 10 \le 0 \mathmr{and} 3 x - 7 \le 0$

Solving for $x$ in each inequality, we find

$x \ge - 5 \mathmr{and} x \ge - \frac{7}{3}$
or
$x \le - 5 \mathmr{and} x \le - \frac{7}{3}$

Because $- \frac{7}{3} > - 5$, we have that $x \ge - \frac{7}{3}$ implies $x \ge - 5$ already. Similarly, $x \le - 5$ implies $x \le - \frac{7}{3}$. Thus, we can rewrite the pairs of inequalities as single inequalities:

$x \ge - \frac{7}{3} \mathmr{and} x \le - 5$

Using interval notation, we can express the set of values which act as solutions for each inequality (note that $\in$ means "is an element of" or "is in":)

$x \ge - \frac{7}{3} \iff x \in \left[- \frac{7}{3} , \infty\right)$
($x$ is in the interval containing all real numbers from and including $- \frac{7}{3}$ to infinity)

$x \le - 5 \iff x \in \left(- \infty , - 5\right]$
($x$ is in the interval containing all real numbers from negative infinity to, and including, $- 5$)

The $\cup$, or "union" symbol allows us to treat multiple sets as a single set. If $A$ and $B$ are sets, then $A \cup B$ is the set which contains all the elements of $A$ and all the elements of $B$. Thus, we can write our entire solution set as a single set by using $\cup$ to combine the intervals.

$x \in \left(- \infty , - 5\right] \cup \left[- \frac{7}{3} , \infty\right)$