Question

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

Answer 1

Answer:

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

Step-by-step 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}