Finding values of $t$ I have this equation - 
$3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0 $ 
I'm struggling on how to find the values of $t$ 
First, I power both sides by $2$ to make the power be a whole number . 
I get 
$3t - \frac{2}{5} t^{-3} = 0 $ 
From here I'm stunned and stuck . Can I get a hint ! Thanks ! 
 A: I don't see how you get the second equation: the square of
$3t^{1/2}-\frac25t^{-3/2}$ is not
$3t-\frac25t^{-3}$. Your original equation is
$$3t^{1/2}-\frac{2}{5t^{3/2}}=0.$$
Putting this over a common denominator gives
$$\frac{15t^2-2}{5t^{3/2}}=0.$$
Can you see where to go now?
A: You can't square.
$(a + b)^2 \ne a^2 + b^2$ but instead equals $a^2 + 2ab + b^2$ so that won't make things simpler.
Also squaring both both sides of an equation will add extraneous solutions.  For example, the problem $x + 1 = 2$ has one solution: $x = 1$.  But if I chose to square both sides $(x+1)^2 = 4^2$ then $x^2 + 2x + 1 = 4$ and $x^2 + 2x - 3 = 0$ so $(x+3)(x-1) = 0$ so $x = 1$ or $x = -3$.  So somehow we now have two solutions.  Where did $x = -3$ come from and is it a valid solution?
Notice when we say $a = b$ there is only one possibility: $a$ and $b$ are the same thing.  But if we square both sides to get $a^2 = b^2$ there are two possibilities:  $a$ and $b$ are the same thing, or $a$ and $b$ are negatives of each other.  
So $x + 1 = 2$ means $x + 1 $ and $2$ are the same thing.  But $(x+1)^2 = 2^2$ means either $x + 1$ and $2$ are the same thing, OR $x+1$ and $-2$ are the same thing.  If $x+1 = -2$ we get ..... $x =-3$... which was the new answer that came out of nowhere.  It is not a valid answer because $x + 1 \ne -2$.  We call that an extraneous solution and this often (usually) occurs if we square both sides.
Squaring both sides isn't wrong but we must test for extraneous solutions.
$3t^{\frac 12} - \frac 25 t^{-\frac 32} = 0$
$(3t^{\frac 12} - \frac 25 t^{-\frac 32})^2 = 0$
$9t - 2*3t^{\frac 12}*\frac 25 t^{-\frac 32} + \frac {4}{25}t^{-3} = 0$
$9t - \frac {12}{5t} + \frac {4}{25t^3} = 0$
And ... that's really not any easier... and it wil have extraneous solutions so ... let's not do it.
If getting rid of the fractional powers are *really * a concern (why are they?) then to get rid of them simply do substitution.
Let $r = t^{\frac 12}$.
Then 
$3t^{\frac 12} - \frac 25 t^{-\frac 32} = 0 \implies$
$3r - \frac 25 r^{-3} = 0\implies$
$(3r - \frac 25 r^{-3})5r^3 = 0 *5r^3 \implies$
$15r^4 - 2 = 0\implies$
$r^4 = \frac 2{15}\implies$
$r = \pm \sqrt[4]{\frac 2{15}}\implies$
$t^{\frac 12} = \pm \sqrt[4]{\frac 2{15}}$.  But $t^{\frac 12} \ge 0$ so
$t^{\frac 12} = \sqrt[4]{\frac 2{15}}\implies$
$t = \sqrt{\frac 2{15}}$.
But there really isn't any reason to be afraid of the half powers.
Simply put everything over a common denominator:
$3t^{\frac 12} - \frac 25t^{-\frac 32} = 0 \implies$
$\frac {15t^2 - 2}{5t^{\frac 32}} = 0\implies$
$15t^2 - 2 = 0\implies$
$t = \pm \sqrt {\frac {2}{15}}$.
But as $t^{1/2}$ implies $t$ is non-negative.
$t = \sqrt{\frac {2}{15}}$.
A: The equation
$$
3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0
$$
is equivalent to
$$
t^{\frac{1}{2}} = \frac{2}{15} t^{\frac{-3}{2}} \;.
$$
Since $t$ cannot be zero$\color{green}{^*}$, you are allowed to multiply both sides by $t^{\frac{3}{2}}$:
$$
t^{2} = \frac{2}{15} \;.
$$
So, the solution is
$$
t = \pm\sqrt{\frac{2}{15}} \;.
$$
$\small\color{green}{^*\text{: If $t$ were zero, we would have $0^{\frac{-3}{2}} = \dfrac{1}{0}$.}}$
