What is the solution set of the quadratic inequality x^2+x-2>0?

Question

asked Oct 15, 2019 229k views

2 Answers

Steve Robbins · Answer 1 · 2019-10-19T17:11:52+0000

Answer:

The solution set is { x | x
$\leq$ -2 or x
$\geq$ 1}

Explanation:

The given inequality is

x^(2) +x-2>0

Let us factor
x^(2) +x-2

so we have

(x+2)(x-1)>0

Let us find zeros of
(x+2)(x-1)

(x+2)(x-1)=0

x+2=0 or
x-1 =0

x= -2 or
x=1

so we have intervals (-∞ , -2) , (-2 , 1) and (1, ∞)

we need to find in which interval is
x^(2) +x-2 is greater than 0

so we will assume the value of x in each interval and will plug it in
x^(2) +x-2 and will check if we get negative or positive value

Let us check the sign of
x^(2) +x-2 in (-∞ , -2)

we can take x=-3 and plug it in
x^(2) +x-2

so we have

(-3)^(2) +(-3)-2= 9-3-2= 4 ( which is greater than 0)

This shows (-∞, -2) is one of the solution set

similarly we can check the sign of
x^(2) +x-2 in (-2,1)

we take x= 0 , so we have

0^(2) +0-2=-2 ( which is less than 0)

This shows (-2,1) is not the solution set

now we check the sign of
x^(2) +x-2 in (1 ,∞)

we can assume x= 2, so we have

2^(2) +2-2 = 4 ( which is greater than 0)

This shows (1 ,∞) is the solution set

Hence the solution set in interval notation (∞ ,-2)∪(1,∞)

we can write this as { x | x
$\leq$ -2 or x
$\geq$ 1}