real values for x

**david18** · January 5th 2008, 07:14 AM

Hi, I have this part of a question that gets me:
I did all the previous steps and got to an equation that must =0:

x^2(-x+12)=0

I know that making x=12 is one value, but for the x^2 part, do i just ignore it?
In my answer book it says x=12 and nothing else, so I'm presuming its something to do with real values of x (even though I don't really know what a 'real' value of x is).

Thanks

**wingless** · January 5th 2008, 07:27 AM

For $x \neq 0$ , divide both sides by $x^2$ .

$\frac{x^2(12-x)}{x^2} = \frac{0}{x^2}$

$12 - x = 0$

$x = 12$ .

We found $x=12$ , which makes the equation true, but don't forget that we didn't test $x$ for $0$ . If we do,
$0 (12 - 0) = 0$

Yes, $x = 0$ is a root too.

$x = \{0,12\}$

**wingless** · January 5th 2008, 07:34 AM

Another way (which you may find easier) to solve the equation:

If the product of two values is $0$ , then at least one of them must be $0$ .

$\underbrace{(x^2)}_{x^2=0}\underbrace{(12-x)}_{12-x=0} = 0$

Either $x^2$ or $12 - x$ or both of them must be $0$ . So,

$x^2 = 0$
$x = 0$

And,

$12-x = 0$
$x = 12$

Thus,

$x = \{0, 12\}$

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