Rational Exponents

**A Beautiful Mind** · July 27th 2009, 01:29 PM

I don't understand this.
$<br /> x^3/2-x^1/2 = 0$

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.

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

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

**SENTINEL4** · July 27th 2009, 01:41 PM

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

**A Beautiful Mind** · July 27th 2009, 01:42 PM

yes it is

**SENTINEL4** · July 27th 2009, 01:49 PM

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

**e^(i*pi)** · July 27th 2009, 01:51 PM

Originally Posted by A Beautiful Mind

yes it is

You can only subtract when the powers divide. If you had $x^{\frac{3}{2}} \div x^{\frac{1}{2}}$ then you could subtract the exponents.

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

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

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

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

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

**ANDS!** · July 27th 2009, 01:58 PM

Originally Posted by A Beautiful Mind

I don't understand this.
$<br /> x^3/2-x^1/2 = 0$

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.

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

...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.

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