Question

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

Answer 1

Answer is A.

The values of x must be greater than or equal to 1 and the value of x must be less than or equal to -2.

x >= 1.....x <= -2

Please see attachment!

Answer 2

Answer: The solution set is $(-\infty, -2)\cup (1, \infty).$

Step-by-step explanation:  We are given to find the solution set of the following quadratic inequality:

$x^2+x-2>0.$

The solution is as follows:

$x^2+x-2>0\\\\\Rightarrow x^2+2x-x-2>0\\\\\Rightarrow x(x+2)-1(x+2)>0\\\\\Rightarrow (x+2)(x-1)>0.$

We know that if 'a' and 'b' are two numbers such that a × b > 0, then either both 'a' and 'b' are greater than 0 or both of them are less than 0.

Therefore, we have

either

$x+2>0,~~~x-1>0\\\\\Rightarrow x>-2,~~~~x>1\\\\\Rightarrow x>1,$

or

$x+2<0,~~~x-1<0\\\\\Rightarrow x<-2,~~~~x<1\\\\\Rightarrow x<-2,$

Thus, the required solution set is x < -2 and x > 1. We can also write the solution set in terms of intervals as $(-\infty, -2)\cup (1, \infty).$