Simple Algebra Problem

1. Sep 6, 2011

process91

1. The problem statement, all variables and given/known data
Solve $u-\sqrt{u^2-1}=2$ for u.

2. Relevant equations

3. The attempt at a solution
I know that there are no real solutions, but I don't understand why the following does not work:

\begin{align}\sqrt{u^2-1}&=u-2\\ u^2-1&=u^2-4u+4\\ u&=\tfrac{5}{4}\end{align}

The second line does not seem to be valid, where I square both sides, but why?

2. Sep 6, 2011

process91

AH, I think I understand now. Just will post here to see if someone else can verify my conclusion:

In most algebraic operations, the statement being made is an if-and-only-if, thus when we say

$$\begin{array}{}4u&=5\\u&=\tfrac{5}{4}\end{array}$$

What we're really saying is $4u=5 \iff u=\tfrac{5}{4}$, which is why we know that $\frac{5}{4}$ is a solution without going back and verifying it (it is implied in the IFF relationship).

On the other hand, with squaring both sides,
$$\sqrt{u^2-1}=u-2 \implies u^2-1=4-4u+u^2$$
however this is not an "iff" implication, since negating either side alone would also produce the same squared equation, and so we need to go back and verify the solution. To put it another way, a $u$ which satisfies the original equation will necessarily satisfy the squared equation, however it is not sufficient. Thus if we find a u which does satisfy the squared equation but it does not satisfy the original, no such u exists.

Obviously any other even-powers are also strictly if relationships. Are there any other common operations to watch out for like this?

3. Sep 6, 2011

gb7nash

It all depends on the sign of u - 2 after you solve for u. If u - 2 is positive when you plug in the possible value, then you obtained a real solution for u. If u - 2 is negative, then this basically states that the principal squareroot is a negative value, so the value is rubbish.

4. Sep 7, 2011

HallsofIvy

Staff Emeritus
When ever you square both sides of an equation or multiply both sides of an equation there is a danger of "introducting" false solutions- that is, numbers that satisfy the new equation but not the original.

For example, the only root of x= 2 is obviously 2. But $x^2= 4$ has both 2 and -2 as roots. Or, the only root to x- 2= 0 is 2 but $(x- 2)(x- 3)= x^2- 5x+ 6= 0$ has both x= 2 and x= 3 as roots.

5. Sep 7, 2011

process91

Thanks guys, that helped clear it up for me.