Here's the question you clicked on:
sabika13
Solve (4-x^2)(x^2-2x+2)<0
(4-x^2)(x^2-2x+2)<0 Let's study the equation. If (4 - x^2) = 0, then 0 < 0 which isn't true. If (x^2 - 2x +2 ) = 0, then 0 < 0 which also isn't true. Now, let's look at (x^2 - 2x +2 ) Let's rewrite this equation (x-1)^2 + 1 If you look at this equation, no matter what value of x you put in, the square will always make it positive (excluding complex numbers but I hope you don't have to deal with that here) What does that mean? It means that (x^2-2x+2)>0 and is a positive number. That means that (4-x^2) must be less than 0. It must be a negative number. How do we figure that out? We solve this: 4-x^2 < 0 If you graph this, you'll see that it's a parabola with it's vertex at (0,4) and it opens downwards. The line of the graph should be dashed which represents the points in this equation does not include the points on that line.