# How to rearrange this square root so a denominator is out the front?

I have the equation:

$$K_f \sqrt{\left(\frac{F_{max}\cdot L\cdot y}{t\cdot I_u}\right)^2 +3\left(\frac{V_{max}}{t\cdot L}\right)^2}$$

And I need to bring the $\mathbf t$ out the front of the square root along with the $\mathbf K_f$ however I don't know what I need to do the equation to do this.

I have tried to make bring the $\mathbf t$ up to the numerator like so:

$$K_f \sqrt{\left(\frac{F_{max}\cdot L\cdot y\cdot t^{-1}}{I_u}\right)^2 +3\left(\frac{V_{max}\cdot t^{-1}}{L}\right)^2}$$

Then I have tried separate it from its quotient:

$$K_f \sqrt{\left(\frac{F_{max}\cdot L\cdot y}{I_u} \cdot \frac{t^{-1}}{1}\right)^2 +3\left(\frac{V_{max}}{L} \cdot \frac{t^{-1}}{1}\right)^2}$$

This is where I am unsure. I have assumed that the product of two squares is the same as the square of the two products:

$$K_f \sqrt{\left(\frac{F_{max}\cdot L\cdot y}{I_u}\right)^2\left(\frac{t^{-1}}{1}\right)^2 +3\left(\frac{V_{max}}{L}\right)^2\left(\frac{t^{-1}}{1}\right)^2}$$

Now I am not sure what to do with the terms with the $\mathbf t$ in it. Would someone be able to help me out in getting both the $\mathbf K_f$ and $\mathbf t$ outside of the square root?

• Your assumption is correct: the product of two squares is the same as the square of the product. Even more appropriately in this case, the square root of a product is the product of the square roots of the factors. The additional piece that you need is that $t^{-2}$ is a common factor of the terms of the sum, so you can use the distributive law to pull it out as one factor of the product - then you can apply the square root to each factor. – NickD Sep 4 '18 at 12:21

You have an expression of the form $$K\sqrt{\left(\frac{A}t\right)^2+\left(\frac{B}t\right)^2}.$$
You can expand the squares, then take the $t$ as common denominator, and then use that the square root of a product is the product of the square roots:
$$K\sqrt{\left(\frac{A}t\right)^2+\left(\frac{B}t\right)^2}=K\sqrt{\frac{A^2}{t^2}+\frac{B^2}{t^2}}=K\sqrt{\frac{A^2+B^2}{t^2}}=K\frac{\sqrt{A^2+B^2}}{\sqrt{t^2}}=K\frac{\sqrt{A^2+B^2}}t=\frac{K}t\sqrt{A^2+B^2}.$$
Here I'm suppossing that $t\geq0$, so that $\sqrt{t^2}=|t|=t$.
In both terms, $t$ appears at the denominator, which is squared. If you pull it out of the square root, the square is just cancelled and $t$ remains at the denominator $(-2\cdot\dfrac12=-1$).
$$K_f \sqrt{\left(\frac{F_{max}\cdot L\cdot y}{t\cdot I_u}\right)^2 +3\left(\frac{V_{max}}{t\cdot L}\right)^2}=\frac{K_f}t \sqrt{\left(\frac{F_{max}\cdot L\cdot y}{ I_u}\right)^2 +3\left(\frac{V_{max}}{L}\right)^2}.$$