# How do you solve  x^2>3x + 4?

Oct 16, 2015

$\left(- \infty , - 1\right) \cup \left(4 , + \infty\right)$

#### Explanation:

First, transfer everything to one side.

$\left[1\right] \textcolor{w h i t e}{X X} {x}^{2} > 3 x + 4$

$\left[2\right] \textcolor{w h i t e}{X X} {x}^{2} - 3 x - 4 > 0$

$\left[3\right] \textcolor{w h i t e}{X X} \left(x - 4\right) \left(x + 1\right) > 0$

Finally, we will create a table of signs. We will use the critical points 4 and -1 (taken from step 3).

Critical Points: -1, 4

$\textcolor{w h i t e}{X X X X X} \left(- \infty , - 1\right) \textcolor{w h i t e}{X} \left(- 1 , 4\right) \textcolor{w h i t e}{X} \left(4 , + \infty\right)$
$\left(x - 4\right) \textcolor{w h i t e}{X X X X} - \textcolor{w h i t e}{X X X X X} - \textcolor{w h i t e}{X X X} +$
$\left(x + 1\right) \textcolor{w h i t e}{X X X X} - \textcolor{w h i t e}{X X X X X} + \textcolor{w h i t e}{X X X} +$
$P r o \mathrm{du} c t \textcolor{w h i t e}{X X X X} + \textcolor{w h i t e}{X X X X X} - \textcolor{w h i t e}{X X X} +$

Since we want the product to be greater than 0, we will choose the intervals where the product is positive (+).

The solution set is the interval:

$\left(- \infty , - 1\right) \cup \left(4 , + \infty\right)$

Oct 16, 2015

Solve x^2 > 3x + 4.

Ans: (-infinity, - 1) and (4, infinity)

#### Explanation:

Write the quadratic inequality in standard form: f(x) = x^2 - 3x - 4 > 0
First, solve y = 0 to find the 2 x-intercepts (real roots).
Factor pairs of (4) --> (-1, 4). This sum is 3 = -b. Then, the 2 real roots are: -1 and 3.
The parabola opens upward (a > 0). Between the two x-intercepts, f(x) is negative. f(x) > 0 outside the interval (-1, 4)
Answer. Open intervals: (-infinity, -1) and (4, infinity).

graph{x^2 - 3x - 4 [-10, 10, -5, 5]}

NOTE. The method that discusses the signs of the 2 binomials is unadvised because it is complicated and easily leads to errors/mistakes.