Why Does Simplifying an Irrational Equation Factor Lead to an Impossible Result?

[greg_rack]

Homework Statement

$$a-2=-\sqrt{a^2-4}$$

Relevant Equations

none

For solving this equation I must take elevate to the square of each member, resulting in:
$$(a-2)^2=a^2-4 \rightarrow a=2$$
Now, the thing I noticed and don't get is that if you simplify a $(a-2)$ factor, the equation becomes impossible:
$$a-2=a+2$$
It must be a stupid thing which I'm missing...

 

[etotheipi]

You divided by $(a-2)$, which is zero!

 

[greg_rack]

You divided by $(a-2)$, which is zero!

Told you it would have been stupid... but how could I know in advance $a-2$ to be 0 if I still haven't found $a$?

 

[etotheipi]

In general, you shouldn't divide by a term unless you have verified that it is not zero. You need to do it the long way!

 

[Mark44]

Homework Statement:: $$a-2=-\sqrt{a^2-4}$$
Relevant Equations:: none

but how could I know in advance a−2 to be 0 if I still haven't found a?

The first thing to notice here is that since $-\sqrt{a^2 - 4}$ is negative less than or equal to 0, then $a - 2$ also has to be negative less than or equal to 0. This implies that a < 2 $a \le 2$.
Also, for the square root to be real, we must have $a^2 - 4 \ge 0$. This implies that $a \ge 2$ or $a \le -2$.
From these restrictions we see that there is no real solution possible only one solution possible.

For solving this equation I must take elevate to the square of each member, resulting in:
$$(a-2)^2=a^2-4 \rightarrow a=2$$
Now, the thing I noticed and don't get is that if you simplify a $(a-2)$ factor, the equation becomes impossible:
$$a-2=a+2$$
It must be a stupid thing which I'm missing...

 

[etotheipi]

The first thing to notice here is that since $-\sqrt{a^2 - 4}$ is negative, then $a - 2$ also has to be negative. This implies that a < 2.
Also, for the square root to be real, we must have $a^2 - 4 \ge 0$. This implies that $a \ge 2$ or $a \le -2$.
From these restrictions we see that there is no real solution possible.

The only restriction on $-\sqrt{a^2-4}$ is $-\sqrt{a^2 - 4} \leq 0$, so $a=2$ is a real solution.

 

[Mark44]

My restriction was a tad too restrictive, and disallowed the possibility that $-\sqrt{a^2 - 4}$ could be zero.
The revised restrictions are $a - 2 \le 0 \Rightarrow a \le 2$ and $a \ge 2$ or $a \le -2$. So $a = 2$ is the only possible solution.
I've edited my earlier post.

 

[Office_Shredder]

As far as how do you know you're dividing by zero, anytime you divide by anything you can just say "either the new equation is true, or the thing I divided by is zero" and then handle the two cases separately.