Thread: How to solve x^2 + 2x - 8 > 0 algebraically

But what to do about the -3. Eg Do I say:

or

It seems that that it has to be x + 1 < -3 - works if look at curve graphically.

But then what if problem is



and which is wrong. Should be
and .
-4 < x < 2 - can see graphically that is under x axis and so between -4 and 2.

So what is the rule for bit when get square root of either side???

Originally Posted by angypangy

But what to do about the -3. Eg Do I say:

or

It seems that that it has to be x + 1 < -3 - works if look at curve graphically.

But then what if problem is



and which is wrong. Should be
and .
-4 < x < 2 - can see graphically that is under x axis and so between -4 and 2.

So what is the rule for bit when get square root of either side???

You need to say that

Then either

Hello, angypangy!

. . . . Not quite

When we take the square root, we get: .

This means: .

The solution is: .


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Graphically, we have a parabola: .

The question is: when is the graph above the x-axis?

Since the parabola opens upward,
. . it is positive outside of its x-intercepts.

The x-intercepts are: .

And we can "see" the solution . . .

Code:

                |
     ♥          |      ♥
                |
                |
      ♥         |     ♥
                |
       ♥        |    ♥
    - - * - - - + - * - - -
       -4 *     | * 2
              * |
                |

If the problem were: .
. . the solution is between the x-intercepts.
. . . . .