# How do you solve x^2+5x+6>0?

Mar 22, 2018

${x}_{1} > - 2 \mathmr{and} {x}_{2} < - 3$

#### Explanation:

${x}^{2} + 5 x + 6 > 0$
$\textcolor{b l u e}{{x}^{2} + 2 \cdot \frac{5}{2} x + {\left(\frac{5}{2}\right)}^{2}} - {\left(\frac{5}{2}\right)}^{2} + 6 > 0$
$\textcolor{b l u e}{{\left(x + \frac{5}{2}\right)}^{2}} - \frac{25}{4} + 6 > 0$
${\left(x + \frac{5}{2}\right)}^{2} - \frac{1}{4} > 0 | + \frac{1}{4}$
${\left(x + \frac{5}{2}\right)}^{2} > \frac{1}{4} | \sqrt{}$
$x + \frac{5}{2} > \pm \frac{1}{2} | - \frac{5}{2}$

${x}_{1} > - 2 \mathmr{and} {x}_{2} < - 3$

If you struggle to understand any of the steps I made,
feel free to write a comment :)

Mar 22, 2018

Range satisfying the given condition: color(blue)(x<(-3) or x >(-2)

We can also write the solution using the interval notation as:

color(blue)((-oo,-3) uu (-2, oo)

#### Explanation:

Given:

We are given the inequality:

color(red)(x^2+5x+6>0

$\textcolor{p u r p \le}{\text{Step 1}}$

Write this inequality as color(green)(x^2+5x+6=0 to factorize.

Consider the quadratic expression ${x}^{2} + 5 x + 6$

Split the middle term to factorize as shown below:

We want two numbers that multiply together to make 6, and add up to 5.

${x}^{2} + 3 x + 2 x + 6$

Factor the first two terms and the last two terms separately:

$x \left(x + 3\right) + 2 \left(x + 3\right)$

Observe that $\left(x + 3\right)$ is a common factor to both the terms.

Hence, we can write our factors as color(blue)((x+3)(x+2)

$\textcolor{p u r p \le}{\text{Step 2}}$

We will construct the sign chart:

$\left(i\right)$ Compute the signs of $\left(x + 2\right) :$

$x + 2 = 0$

Add color(red)(-2 to both sides of the equation.

x+2+color(red)((-2))=0+color(red)((-2)

x+cancel 2+color(red)((-cancel 2))=0+color(red)((-2)

$x = - 2$

Hence, $x + 2$ is ZERO for $x = - 2$

Similarly, $\left(x + 2\right)$ is negative for $x < - 2$

$\left(x + 2\right)$ is positive for $x > - 2$

$\textcolor{p u r p \le}{\text{Step 3}}$

$\left(i i\right)$ Compute the signs of $\left(x + 3\right) :$

$x + 3$ is ZERO for $x = - 3$

$x + 3$ is negative for $x < \left(- 3\right)$

$x + 3$ is positive for $x > \left(- 3\right)$

We will summarize and create a table of values.

Hence,

Range satisfying the given condition: color(blue)(x<(-3) or x >(-2)

We can also write the solution using the interval notation as:

color(blue)((-oo,-3) uu (-2, oo)

An image of the graph for the inequality is available below:

Hope it helps.