# How do you solve x^2 + 7x + 6 <= 6?

Oct 5, 2016

Solution is $- 7 \le x \le 0$

#### Explanation:

As ${x}^{2} + 7 x + 6 \le 6$, we have ${x}^{2} + 7 x + \cancel{6} \le \cancel{6}$ or

${x}^{2} + 7 x \le 0$ or $x \left(x + 7\right) \le 0$

If product of $x$ and $x + 7$ is negative (for the time we are ignoring equality sign - note that for equality we just add $x = 0$ and $x = - 7$ to the solution),

either $x > 0$ and $x + 7 < 0$ i.e. $x < - 7$ - but $x > 0$ and $x < - 7$ together is just not possible.

or $x < 0$ and $x + 7 > 0$, i.e. $x \succ 7$ - which is possible if $x$ lies between $- 7$ and $0$ i.e. $- 7 < x < 0$.

Now including the equality sign

Solution is $- 7 \le x \le 0$.

You may also see from the graph that ${x}^{2} + 7 x$ is negative in the interval $- 7 \le x \le 0$.
graph{x(x+7) [-10, 5, -20, 20]}