How to solve that equation? Do I use quadratic equation?

tomyy2012 said:
Yes I would like to calculate the value of Q
Click to expand...
First bring all the terms to one side of the equation and simplify the given equation. Then apply the quadratic equation.

What does your textbook/notes say about solution to quadratic equation?
 
By inspection, you can see this will be a quadric. So aim to arrange your equation in the form:
AQ^2 + BQ + CQ = 0
Now see if it can be factored easily? Probably not worth a lot of effort on that so go straight to quadric solving equation.
This will give youTWO possible solutions of Q in terms of A,B,C

hint: multiplying everything by the same number does not change the solution for example 0.000006 * 1000000 = 6
 
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Yes, that's a quadratic equation because it has a polynomial in Q with highest power 2. Seeing that there are "-0.012Q" on one side of the equation and "-0.008Q" on the other, I would combine them by adding 0.008Q to both sides:
\(\displaystyle 8-0.012Q+ 0.008Q- 0.000006Q^2= 10- 0.008Q+ 0.008Q\)
\(\displaystyle 8-0.004Q- 0.000006Q^2= 10\).

There are constants on both sides so combine them- subtract 10 from both sides:
\(\displaystyle 8- 10- 0.004Q- 0.000006Q^2= 10- 10\).
\(\displaystyle -2- 0.004Q- 0.000006Q^2= 0\).

Now, you could put those numbers into the "quadratic formula but I would be inclined to firt divide throught by "-0.000006" to try to make the numbers "nicer".

\(\displaystyle \frac{-2}{-0.000006}- \frac{0.004}{-0.000006}- \frac{0.000006}{-0.000006}Q^2= 0\)
\(\displaystyle Q^2+ 666\frac{2}{3}Q+ 333\frac{1}{3}= 0\).

Looking at that, I think I would "back up" a bit by multiplying by 3:
\(\displaystyle 3Q^2+ 2000Q+ 1000= 0\).

Now use the "quadratic formula":
\(\displaystyle Q= \frac{-2000\pm\sqrt{(2000^2)- 4(3)(1000)}}{6}\).
 
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