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This is the problem I have:

$$y^{\frac{3}{2}} = 5y$$

What I tried so far was raising $y^{\frac{3}{2}}$ to the $\frac{3}{2}$, making it equal $1$, but I had trouble raising $5y$ to that power.

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2 Answers 2

up vote 2 down vote accepted

You have the right idea, you are trying to get rid of the fractional power. However, the best way to do this is to raise each side to the power $2$, so we have:

$$\left(y^{\frac{3}{2}}\right)^{2}=\left(5y\right)^{2}\implies y^{3}=25y^{2},$$

Which is then much easier to manipulate.

Hope this helps!


If you are trying to find all solutions to the equation, then you must solve a cubic equation:

$$y^{3}-25y^{2}=0$$

This can be factorized as:

$$y^{2}\left(y-25\right)=0,$$

Which gives us the following solutions:

$$y=\{0,25\}$$

Where $0$ is a duplicated root.

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Ah, yes. But how would I move on from here? 25 isn't a cube. –  Andre Oseguera Aug 21 '12 at 14:22
    
Of course this does help him, Shaktal. –  Babak S. Aug 21 '12 at 14:22
    
@AndreOseguera Are you trying to find all solutions to the equation? –  Shaktal Aug 21 '12 at 14:22

I probably would first write the equation $y^{3/2}=5y$ as $y(y^{1/2})=5y$. This has the solution $y=0$. If $y \ne 0$, we can cancel $y$ from both sides (that is, divide both sides by $y$) and obtain $y^{1/2}=5$. Now square both sides. We get $y=25$.

So the two solutions of the equation are $y=0$ and $y=25$.

Remark: Let us explore your idea. You wanted to get a $y$ on the left-hand side by raising $y^{3/2}$ to a certain power. That can be done, but the appropriate power is $2/3$.

We have $(y^{3/2})^{2/3}=y^{(3/2)(2/3)}=y^1=y$. We need to do the same thing to the right-hand side, and we obtain the equation $y=5^{2/3}y^{2/3}$. Improved the left-hand side, but the right-hand side has been uglified.

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