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Given a market demand function:

$$p=4-q-q^2$$

and the market supply function:

$$p=1+4q+q^2$$

We can solve for:

$$4-q-q^2 = 1+4q+q^2$$

When I move the terms from right to left, I get $3-5q-2q^2$, but from left to right $2q^2+5q-3$.

I understand multiplying by $-1$ will give me the inverse of the other.

To determine the equilibrium price and quantity, we can use the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

However, this will give a different result depending on which direction I have collected the terms.

My question is therefore, how do I know which of the two I use to solve the problem?

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    $\begingroup$ What is your question? $\endgroup$ – user259242 Nov 20 '16 at 11:20
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    $\begingroup$ "Obviously, this gives a different result depending on the direction": sorry but there is nothing obviously different. Solve these two equations to see. $\endgroup$ – Yves Daoust Nov 20 '16 at 11:28
  • $\begingroup$ Sorry, have clarified my question a little. $\endgroup$ – Dan Nov 20 '16 at 11:38
  • $\begingroup$ Note that by definition $a$ is the coefficient of the $q^2$ term, $b$ the coefficient of the $q^1$ term and c the coefficient of the $q^0$, aka the constant term. $\endgroup$ – Hyperplane Nov 20 '16 at 12:09
  • $\begingroup$ Okay, taking the q^2 term we still have 2q^2 and -2q^2, so how would I know which one was correct? $\endgroup$ – Dan Nov 20 '16 at 12:14
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When you move arguments from right to left you get:

$3 - 5q - 2q^2 = 0$

and you can multiply this expression by (-1) and keep the equivalency, so you get:

$2q^2 + 5q - 3 = 0$

When you solve these for x, you will get the same solutions.

Edit: the solutions for x of both equations are 1/2 and -3, you must have missed something while putting the numbers in the formula. (put both these numbers instead of x, you should get an equality.) I can only assume you missed a minus somewhere, because the +/- operator sometimes confuses people.

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  • $\begingroup$ I've update my question a little. Could you shed some light? $\endgroup$ – Dan Nov 20 '16 at 11:59
  • $\begingroup$ I've edited my answer, it should be okay now. :) $\endgroup$ – Collapse Nov 20 '16 at 14:08

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