How Do You Solve for Q in a Complex Algebra Equation?

[jmb07]

how do i go about solving for q in the equation:

(F x r x r)/k = (6.63E-6 -q)(q)

 

[RoyalCat]

What are your knowns/unknowns and are there any additional equations?

I skimmed over your other thread, and you're trying to solve for q where the rest of the variables are known, right?

$$\frac{Fr^2}{k}=(6.63*10^{-6} -q)q$$
This is just a simple quadratic equation. Using factoring here obviously won't be useful since the numbers are a bit messy. Just arrange it to the general form ($$aq^2+bq+c=0$$) and solve for q, and remember any limits you may have on its value because of its physical properties.

 

[Architeuthis Dux]

To solve for q in this equation, you will need to use algebraic principles to isolate the variable q on one side of the equation. First, distribute the parentheses on the right side of the equation to get:

(F x r x r)/k = 6.63E-6q - q^2

Next, bring all terms containing q to one side of the equation by subtracting 6.63E-6q from both sides:

(F x r x r)/k + 6.63E-6q = -q^2

Then, factor out q on the left side of the equation:

q(-q + 6.63E-6) = (F x r x r)/k

To isolate q, divide both sides of the equation by (-q + 6.63E-6):

q = (F x r x r)/k/(-q + 6.63E-6)

Finally, use the quadratic formula to solve for q:

q = [(F x r x r)/k ± sqrt((F x r x r)^2 - 4k(-q + 6.63E-6))]/2k

Note that there may be two possible solutions for q, depending on the values of F, r, and k.