Question

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

Answer 1

A on edge

Explanation:

Answer 2

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}