1. Rational Exponents

I don't understand this.
$\displaystyle x^3/2-x^1/2 = 0$

For the problem, you're suppose to factor out a 1/2. I don't get that. You can't just go ahead and subtract?

Someone please explain this problem to me.

$\displaystyle x^3/2=x^1/2$

...3/2 and 1/2 are the exponents.

2. $\displaystyle x^\frac{3}{2}=x^\frac{1}{2}$ is your problem?

3. yes it is

4. $\displaystyle x^\frac{3}{2}-x^\frac{1}{2}=0 \Rightarrow x^\frac{1}{2}(x-1)=0$
So the solutions are $\displaystyle x=0$ or $\displaystyle x=1$

5. Originally Posted by A Beautiful Mind
yes it is
You can only subtract when the powers divide. If you had $\displaystyle x^{\frac{3}{2}} \div x^{\frac{1}{2}}$ then you could subtract the exponents.

You can also think of it as the definition of exponents:

$\displaystyle a^3 -a^2 = a \times a \times a - a \times a \neq a^{3-2}$

By inspection it would appear only 0 and 1 are solutions and this can be confirmed by factorising:

$\displaystyle x^{\frac{3}{2}} - x^{\frac{1}{2}} = x^{\frac{1}{2}}(x-1)= 0$

Either $\displaystyle \sqrt{x} = 0 \: \rightarrow \: x = 0 \: \text { or }\: (x-1) = 0 \: \rightarrow \: x = 1$

6. Originally Posted by A Beautiful Mind
I don't understand this.
$\displaystyle x^3/2-x^1/2 = 0$

For the problem, you're suppose to factor out a 1/2. I don't get that. You can't just go ahead and subtract?

Someone please explain this problem to me.

$\displaystyle x^3/2=x^1/2$

...3/2 and 1/2 are the exponents.
Well, just factoring isn't going to help you (well it will in this case, because its a pretty obvious question).

You want to first arrange your terms, and then solve the resultant equation:

[X^(3/2)]-[X^(1/2)]=0

Factoring out an X^(1/2) and not just a (1/2) yields:

[X^(1/2)][X^(2/2)-1], which is of course just (X-1). From there it should be fairly easy to figure out what you're answers are.