Determine the set of all real x satisfying
x^2 + 3x >= -2 + x
Enter your answer in interval notation.
+36 You want to have a quadratic one one side that is equal to zero. You can do this by adding 2-x to both sides, leaving you with:
x^2 + 2x + 2 >= 0
You will notice that this is not factorable but it does not need to be factored. You can see that the discrimant, which is the term under the square root of the Quadratic Formula (b^2 - 4ac) is negative. When the Discrimant is negative, it means that the quadratic has no real solutions, or no real x-intercepts. Since the a value is positive, thus meaning that the parabola opens upwards, we know that all values of x result in positive y values. So, the answer to the question is:
(-inf,inf), or, in other words, all values of x will suffice.
Let me know if I did anything wrong!
