Vtyjkyuk

I have the equation:

$$K_f sqrtleft(fracF_maxcdot Lcdot ytcdot I_uright)^2 +3left(fracV_maxtcdot Lright)^2$$

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:

$$K_f sqrtleft(fracF_maxcdot Lcdot ycdot t^-1I_uright)^2 +3left(fracV_maxcdot t^-1Lright)^2$$

Then I have tried separate it from its quotient:

$$K_f sqrtleft(fracF_maxcdot Lcdot yI_u cdot fract^-11right)^2 +3left(fracV_maxL cdot fract^-11right)^2$$

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:

$$K_f sqrtleft(fracF_maxcdot Lcdot yI_uright)^2left(fract^-11right)^2 +3left(fracV_maxLright)^2left(fract^-11right)^2$$

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?

You have an expression of the form
$$Ksqrtleft(fracAtright)^2+left(fracBtright)^2.$$

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:

$$Ksqrtleft(fracAtright)^2+left(fracBtright)^2=KsqrtfracA^2t^2+fracB^2t^2=KsqrtfracA^2+B^2t^2=KfracsqrtA^2+B^2sqrtt^2=KfracsqrtA^2+B^2t=fracKtsqrtA^2+B^2.$$

Here I'm suppossing that $tgeq0$, so that $sqrtt^2=|t|=t$.

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 $(-2cdotdfrac12=-1$).

$$K_f sqrtleft(fracF_maxcdot Lcdot ytcdot I_uright)^2 +3left(fracV_maxtcdot Lright)^2=fracK_ft sqrtleft(fracF_maxcdot Lcdot y I_uright)^2 +3left(fracV_maxLright)^2.$$