# I went awry somewhere with this problem

#### hb7of9

Hi there,

I was expected an answer of 2 for this problem, but it seems the correct answer is 8!

:cold::cold::cold::cold:

Where did I go wrong?

I replaced the m or 4, then went 4-2 which got me to 2.

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#### greg1313

Forum Staff
Problem statement:

Given $g(m)=\sqrt{m-4}$, solve $g(m)=2$.

Solution:

$$\displaystyle 2=\sqrt{m-4} \\ 2^2=(\sqrt{m-4})^2 \\ 4=m-4 \\ 4+4=m \\ 8=m$$

1 person

#### hb7of9

Hi Greg,

thanks so much for the solution: I am super confused though -- must be missing something fundamental in my math knowledge.

For example:
In the second line of the solution you added the exponent to both sides of the equation -- why? How is that the correct thing to do?

Also it feels to me like the root sign disappeared and nothing was done with it --but I am sure that's not the case?

Yikes, I got a lot to learn

#### Greens

In the second line of the solution you added the exponent to both sides of the equation -- why? How is that the correct thing to do?
Here's why that works.

Let $a=b$, then you can multiply by $a$ on both sides to get $a^2 = ab$, or you can multiply by $b$ on both sides to get $ab=b^2$.

Notice that both $a^2 = ab$ and $b^2 = ab$, so we have $a^2 = b^2$

You can generally apply exponents to both sides like you would add/multiply on both sides.

Also it feels to me like the root sign disappeared and nothing was done with it --but I am sure that's not the case?
A square root is actually just an exponent of $\frac{1}{2}$. For example $\sqrt{a} = a^{\frac{1}{2}}$

So you can get rid of the radical by squaring it. $\sqrt{a}^{2} = (a^{\frac{1}{2}})^{2} = a^{\frac{2}{2}} = a^1 = a$

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2 people

#### skipjack

Forum Staff
I replaced the m or 4, then went 4-2 which got me to 2.
Replaced with what and why? What does "went 4-2" mean and why did you do it?

#### DarnItJimImAnEngineer

In the second line of the solution you added the exponent to both sides of the equation -- why? How is that the correct thing to do?
First rule of algebra: Whatever you do to one side of the equation, you have to do to the other.

Philosophy of tackling algebra problems: If you're solving for x, you want to get x on its own on one side of the equation. You generally do this by seeing what was done to x, and then undoing it.*

If you have something under a square root, try squaring it. If you have x + 3, try subtracting 3.

I think Greg and Greens explained the rest of it quite nicely.

*Note, there are other rules and special cases you have to watch out for, but it sounds like you're probably not there yet.