How to solve $x^{1/2}-x^{1/3} = 0$

Question

How can I solve the following equation? I really can't figure out how to solve it:

$x^{1/2}-x^{1/3} = 0$

Thank you.

Answer 1

$$\begin{eqnarray*}x^{1/2}-x^{1/3} &=& 0 \\ \\ \iff x^{3/6} - x^{2/6} &=& 0 \\ \\ \iff x^{2/6}(x^{1/6} - 1)&=& 0 \\ \\ \iff [x^{1/6} = 1\text{ or}\;x^{2/6} &=& 0] \\ \\ \iff [x = 1\text{ or}\;x &=& 0]\end{eqnarray*}$$

Answer 2

Putting $x=y^6, y^3=y^2,y^2(y-1)=0\implies y=0$ or $1\implies x=0$ or $1$

Answer 3

You might have heard this a thousand times, but the equation really does speak here!

What it says is that when you raise a number ($x$) to two different powers ($\frac 12$ and $\frac 13$) they turn out to be equal (their difference is $0$).

The only two numbers that come to mind are $0$ and $1$. $\Big[ \forall n \in \mathbb R, \;\; 0^n=0$ and $1^n=1 \Big]$

So the solution should be $x=0\ \text{ or } \ 1$

Answer 4

(Henceforth, $x$ is a variable denoting a non-negative real number)

$$x^{1/2} - x^{1/3} = 0$$ if and only if $$x^{1/2} = x^{1/3} $$ if and only if $$\frac{x^{1/2}}{x^{1/3}} = 1 \qquad \text{or} \qquad x^{1/3} = 0 $$ if and only if $$x^{\frac{1}{2} - \frac{1}{3}} = 1 \qquad \text{or} \qquad x = 0 $$ if and only if $$x^{1/6} = 1 \qquad \text{or} \qquad x = 0 $$ if and only if $$x = 1 \qquad \text{or} \qquad x = 0 $$

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$$x^{1/2} - x^{1/3} = 0$$ if and only if $$x^{1/2} = x^{1/3} $$ if and only if $$x = x^{2/3} $$ if and only if $$x^3 = x^2 $$ if and only if $$x = 1 \qquad \text{or} \qquad x^2 = 0 $$ if and only if $$x = 1 \qquad \text{or} \qquad x = 0 $$

Answer 5

$x^{1/2}-x^{1/3} = 0$

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

$(x^{1/2})^6=(x^{1/3})^6$

$x^3=x^2$

$x^3-x^2=0$

$x^2(x-1)=0$

$x^2 = 0$ or $x-1=0$

$x = 0$ or $x = 1$

Answer 6

Similar to the first answer.

$$x^{\frac{1}{2}}-x^{\frac{1}{3}}=0$$ Rewrite $x^{\frac{1}{2}}$ as something clever: $$x^{\frac{3}{6}}-x^{\frac{1}{3}}=0$$ (In case that's illegible, that's x^(3/6))

Now factor out $x^{\frac{1}{3}}$: $$x^{\frac{1}{3}}(x^{\frac{2}{6}}-1)=0$$

Now divide both sides by $x^{\frac{1}{3}}$. Note that this assumes $x \neq 0$; we will have to go back and check for that case later.

$$x^{\frac{2}{6}}-1=0$$ $$x^{\frac{2}{6}}=1$$ $$x^{\frac{1}{3}}=1$$ $$(x^{\frac{1}{3}})^3=1^3$$ $$x=1$$

Now let's go back and check if it works for $x=0$:

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

Yep, that equation holds!

So either $x=1$ or $x=0$.

Answer 7

This is late, but for problems like these, I use Wolfram Alpha if I'm stuck. If you log in, it lets you see 3 step-by-step solutions daily.

Here's the page for this problem.