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

• Dec 17th 2011, 05:58 AM
angypangy
How to solve x^2 + 2x - 8 > 0 algebraically
$\displaystyle x^2 + 2x - 8 > 0$

$\displaystyle (x + 1)^2 - 9 > 0$

$\displaystyle (x + 1)^2 > 9$

$\displaystyle x + 1 > 3$ But what to do about the -3. Eg Do I say:

$\displaystyle x + 1 > -3$ or $\displaystyle x + 1 < -3$

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

But then what if problem is $\displaystyle x^2 + 2x - 8 < 0$

$\displaystyle (x + 1)^2 < 9$

$\displaystyle x + 1 < 3$ and $\displaystyle x + 1 < -3$ which is wrong. Should be
$\displaystyle x + 1 < 3$ and $\displaystyle x + 1 > -3$.
-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???
• Dec 17th 2011, 06:30 AM
e^(i*pi)
Re: How to solve x^2 + 2x - 8 > 0 algebraically
Quote:

Originally Posted by angypangy
$\displaystyle x^2 + 2x - 8 > 0$

$\displaystyle (x + 1)^2 - 9 > 0$

$\displaystyle (x + 1)^2 > 9$

$\displaystyle x + 1 > 3$ But what to do about the -3. Eg Do I say:

$\displaystyle x + 1 > -3$ or $\displaystyle x + 1 < -3$

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

But then what if problem is $\displaystyle x^2 + 2x - 8 < 0$

$\displaystyle (x + 1)^2 < 9$

$\displaystyle x + 1 < 3$ and $\displaystyle x + 1 < -3$ which is wrong. Should be
$\displaystyle x + 1 < 3$ and $\displaystyle x + 1 > -3$.
-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 $\displaystyle \pm(x+1) > 3$

Then either $\displaystyle x+1 > 3 \text{ or } -(x+1) > 3 \Leftrightarrow x+1 < -3$
• Dec 17th 2011, 06:50 AM
Soroban
Re: How to solve x^2 + 2x - 8 > 0 algebraically
Hello, angypangy!

Quote:

$\displaystyle x^2 + 2x - 8 \,>\, 0$

$\displaystyle (x + 1)^2 - 9 \,>\, 0$

$\displaystyle (x + 1)^2 \,>\, 9$

$\displaystyle x + 1 \,>\, 3$ . . . . Not quite

When we take the square root, we get: .$\displaystyle |x+1| \:>\:3$

This means: .$\displaystyle \begin{Bmatrix}x+1 \:>\:3 & \Rightarrow & x \:>\:2 \\ & \text{or} \\ x+1 \:<\:\text{-}3 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix}$

The solution is: .$\displaystyle (\text{-}\infty,\,-4)\,\cup\,(2,\,\infty)$

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

Graphically, we have a parabola: .$\displaystyle y \:=\:x^2+2x-8$

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: .$\displaystyle (x+4)(x-2) \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}4,\,2$

And we can "see" the solution . . .
Code:

                |     ♥          |      ♥                 |                 |       ♥        |    ♥                 |       ♥        |    ♥     - - * - - - + - * - - -       -4 *    | * 2               * |                 |
If the problem were: .$\displaystyle x^2 + 2x - 8 \;\;{\color{red}<}\;\;0$
. . the solution is between the x-intercepts.
. . . . . $\displaystyle \text{-}4 \;<\;x\,<\;2$