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

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December 17th 2011, 05:58 AM
angypangy

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

$x^2 + 2x - 8 > 0$

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

$(x + 1)^2 > 9$

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

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

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

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

$(x + 1)^2 < 9$

$x + 1 < 3$ and $x + 1 < -3$ which is wrong. Should be
$x + 1 < 3$ and $x + 1 > -3$ .
-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???
December 17th 2011, 06:30 AM
e^(i*pi)

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

Quote:

Originally Posted by angypangy

$x^2 + 2x - 8 > 0$

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

$(x + 1)^2 > 9$

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

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

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

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

$(x + 1)^2 < 9$

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

Then either $x+1 > 3 \text{ or } -(x+1) > 3 \Leftrightarrow x+1 < -3$
December 17th 2011, 06:50 AM
Soroban

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

Hello, angypangy!

Quote:

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

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

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

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

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

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

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

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

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

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

And we can "see" the solution . . .

Code:

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

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