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

2 Answers
Oct 16, 2015

Answer:

#(-oo,-1)uu(4,+oo)#

Explanation:

First, transfer everything to one side.

#[1]color(white)(XX)x^2>3x+4#

#[2]color(white)(XX)x^2-3x-4>0#

Next, factor the quadratic equation.

#[3]color(white)(XX)(x-4)(x+1)>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

#color(white)(XXXXX)(-oo,-1)color(white)(X)(-1,4)color(white)(X)(4,+oo)#
#(x-4)color(white)(XXXX)-color(white)(XXXXX)-color(white)(XXX)+#
#(x+1)color(white)(XXXX)-color(white)(XXXXX)+color(white)(XXX)+#
#Productcolor(white)(XXXX)+color(white)(XXXXX)-color(white)(XXX)+#

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:

#(-oo,-1)uu(4,+oo)#

Oct 16, 2015

Answer:

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.