# How do you solve x^2+16x+24>6x?

May 25, 2018

$x < - 6$or $x \succ 4$

#### Explanation:

simplifying to ${x}^{2} + 10 x + 24 > 0$

solving ${x}^{2} + 10 x + 24 = 0$ we get
${x}_{1} = - 4$
${x}_{2} = - 6$
so our inequality is equivalent to
$\left(x + 4\right) \left(x + 6\right) > 0$
this gives us
$x \succ 4$ or $x < - 6$

Aug 9, 2018

$x \in \left(- \infty , - 6\right) \cup \left(- 4 , \infty\right)$

#### Explanation:

Given:

${x}^{2} + 16 x + 24 > 6 x$

Subtract $6 x$ from both sides to get:

${x}^{2} + 10 x + 24 > 0$

We can make the left hand side into a perfect square trinomial by adding $1$, so let us add $1$ to both sides to get:

${\left(x + 5\right)}^{2} = {x}^{2} + 10 x + 25 > 1$

Note that this would give equality when ${\left(x + 5\right)}^{2} = 1$, i.e. when $x + 5 = \pm 1$, i.e. when $x = - 6$ or $x = - 4$.

Hence the inequality is achieved when:

$x < - 6 \text{ }$ or $\text{ } x > - 4$

In interval notation, when:

$x \in \left(- \infty , - 6\right) \cup \left(- 4 , \infty\right)$