# Math Help - How would I go about solving this eqn for x?

1. ## How would I go about solving this eqn for x?

I'm in Calculus and the Algebra is killing me. I do practice my Algebra when I get free time on the weekends, but I still have a long way to go. How would I solve this problem for x? I've literally been up all day doing these 3 assignments and I literally maybe finished 50% in a 10hr period.

sqrt(16-x^2)-((x^2)/sqrt(16-x^2)) = 0; solve for x.

The answer was somethin like (+or-) 2sqrt(2).

Can someone give me a step by step on how to solve this, because I encounter these problems involving fractions and sqrts A LOT and they slow me down until I give in.

2. ## Re: How would I go about solving this eqn for x?

The very first thing you do ALWAYS is check out what values of x are impossible. The denominator is sqrt(16-x^2), so -4<x<4. This is quite simple math that just involves adding to both sides, multiplying, and factoring. Can you show us what you did so far?

**Hint** Get rid of the denominator to make the problem clean and simple. Doing that might solve more than one problem in the equation.

3. ## Re: How would I go about solving this eqn for x?

Originally Posted by RogerSmith
I'm in Calculus and the Algebra is killing me. I do practice my Algebra when I get free time on the weekends, but I still have a long way to go. How would I solve this problem for x? I've literally been up all day doing these 3 assignments and I literally maybe finished 50% in a 10hr period.

sqrt(16-x^2)-((x^2)/sqrt(16-x^2)) = 0; solve for x.

...
1. Determine the domain of this equation. You should come out with $x \in (-4, 4)$.

2. $\sqrt{16-x^2} - \frac{x^2}{\sqrt{16-x^2}}=0$ ................ multiply through by $\sqrt{16-x^2}$

which yields:

$16-x^2 - x^2 = 0~\implies~x^2=8~\implies~x=2\sqrt{2} \vee x = -2\sqrt{2}$

4. ## Re: How would I go about solving this eqn for x?

Originally Posted by earboth
1. Determine the domain of this equation. You should come out with $x \in (-4, 4)$.

2. $\sqrt{16-x^2} - \frac{x^2}{\sqrt{16-x^2}}=0$ ................ multiply through by $\sqrt{16-x^2}$

which yields:

$16-x^2 - x^2 = 0~\implies~x^2=8~\implies~x=2\sqrt{2} \vee x = -2\sqrt{2}$
Is there a guide on here that shows me how to write out problems the way that you did?

Also, that is the part that was confusing me in your step 2. I don't know whether to get the fraction over to the right side of the equation and multiply both sides by the reciprocal or did you just look at the fraction and take the reciprocal of the denominator and multiply. That's where I get confused. This is stuff I should've learned in high school but sadly I had to get the teachers who never stayed in class or were never at work.

From now on, I'll upload snapshots of the work I've done so far.

5. ## Re: How would I go about solving this eqn for x?

The first post mentioned these are homework assignments. Please respond accordingly.

-Dan

6. ## Re: How would I go about solving this eqn for x?

Sorry about that guys, I went to sleep. Anyway they gave me another problem to solve because I already had the wrong answer when I came for help. The problem is very similar to the last, the only thing that changed are the numbers. Here is my new problem and the work I did so far. Don't worry about steps a,b,and c at the top. I can do those. I just wanted to know if my algebra is right up to this point. This image may be pretty big. Thanks for all your help guys, I greatly appreciate it.

7. ## Re: How would I go about solving this eqn for x?

Originally Posted by RogerSmith
Sorry about that guys, I went to sleep. Anyway they gave me another problem to solve because I already had the wrong answer when I came for help. The problem is very similar to the last, the only thing that changed are the numbers. Here is my new problem and the work I did so far. Don't worry about steps a,b,and c at the top. I can do those. I just wanted to know if my algebra is right up to this point. This image may be pretty big. Thanks for all your help guys, I greatly appreciate it.
Hello,

Obviously you are not quite certain if the last 2 or 3 lines are correct:

The solution of the equation

$x^2 = c$ , c is a constant

is

$|x|=\sqrt{c}$

Using the definition of the absolute value you'll get

$x = -\sqrt{c}\ \vee \ x = \sqrt{c}$

So you see: Your result is OK, but your method could be a little bit better.