0

I want a solution of the following equation :

$$m^2+n^2-2mn-m=0$$ with the following constraint :

$$0<m<1, 0<n<1,m+n<1,m>n$$

  • 1
    What if there is no solution with these constraints? (Hint: there is none) – Daniel Fischer Sep 27 '13 at 11:03
  • @DanielFischer: How to prove such things ? –  Sep 27 '13 at 11:14
  • 3
    The easy way is to see that the equation is $(m-n)^2 = m$. Since everything is positive by the constraints, and $m > n$, we have $m-n = \sqrt{m}$, hence $m > \sqrt{m}$. But $0 < m < 1$ implies $m < \sqrt{m}$. – Daniel Fischer Sep 27 '13 at 11:19

1 Answers1

0

Hint: You can treat this as a quadratic equation, if you think of $m$ as your variable and $n$ as a constant: $$ \underbrace{m^2}_{\text{quadratic}}+\underbrace{(-2n-1)m}_{\text{linear}}+\underbrace{n^2}_{\text{constant}}=0 $$ In other words, your quadratic is $am^2+bm+c$, where $a=1$, $b=(-2n-1)$, and $c=n^2$. With this, you can come up with two possibilities for how we can write $m$ in terms of $n$. Then, it is a matter of exploring the constraints.

Alternatively:

Since we require that $m>n$, we know that $m-n>0$. We also know that $m>0$. Our equation can be rewritten as $(m-n)^2-m=0$, or $(m-n)^2=m$; since $m>0$, we can take square roots of both sides to get $$ \lvert m-n\rvert=\sqrt{m}. $$ What simplifications can be made here based on the assumption that $m-n>0$? What does this lead you to?

Nick Peterson
  • 32,430