Question: Quadratic Inequality with x

If we end up picking the negative region first correctly and get a negative result, is there any reason to test the other regions? Thanks.

Is it always the middle region that gives the negative answer?

Isn't it an overkill to solve for the values of x and still have to plot these values on the number line to determine the range of values that satisfy the inequality? We know that the value of x that will make the expression to be negative (less than 0) has to fall withing this range of values (in this case, 1 and 3. so, 1<x<3 and since we are looking for an integer, x has to be 2.

Unless you can show me an example where the value of x is outside the range of the solution resulting from setting the expression to zero, I have to say that it is a total waste of time graphing the solution on the number line. That's just my take.

In addition to what you've said, what helped it make sense in my mind, was to think about the problem graphically. All quadratic equations are parabolas. This one is facing upward because x^2 is positive. Solving for when y = 0 gives us the x intercepts. So, we know between the x intercepts the values will be negative and to the right and left of the x intercepts the values will be positive.

If we had a negative value in front of x, like -x2 − 8x − 7, things would be opposite, as our parabola would open down. The region between the y=0 values would be positive, and outside of the y=0 areas would be negative.

I presume we can also say that since the inequality is <0 in this case that we know that the bottom of the parabola will fall below the y axis.

If you have a cubic equation or higher, then I guess you do have to test each region.

So, looking at your explanation and the mathisfun one together it all seems to make sense.

I guess my point is it might help people if they remember that a quadratic equation will always be a parabola and that this problem can also be thought of graphically. (Of course maybe everyone else does keep that in mind).

I didn't use number line to solve the problem, instead I took this route:
--> x^2 - 4x + 3 + 1 - 1 < 0
--> (x^2 - 4x + 4) - 1 < 0
--> (x-2)^2 - 1 < 0
--> (x-2)^2 < 1
The only way for this equation to be less than one is when x = 2

Hi Brent, so we found out is negative between 1 & 3. But then How is 2 = negative? Isn't this contradicting? Could you help clarify? Thanks Brent