# Thread: I need help-code and math

1. ## Re: I need help-code and math

It’s been thirty years since I have had to square root anything. I had to look it up. When a number is multiplied by itself is the answer.

I get that part, it's just like coding.

Originally Posted by grapes
Take the square root of both sides.
I would do this to get rid of the squared on the left to simplify the equation and that would make the 9 on the right 3.

Originally Posted by grapes
The square root of the left hand side is easy, the square root is the inverse of the squaring, so all we're left with is (3x-5), with two possibilities, that it could be the positive or the negative square root.
Negative and positive possibilities could happen in other equations, and that is why both exist, right?

2. ## Re: I need help-code and math

The point is, what we do to the equation can affect the solution, and we sometimes have to account for those changes.

For instance, x=3 is a simple equation that has a simple solution. There is only one value for x that makes it true, and that value is 3. If we were to square both sides, we get which has two solutions, 3 and -3. By squaring both sides, we've introduced the possibility of a solution that wasn't a solution to the original. Similarly, when we take the square root of both sides, we have to remember to keep both solutions.

That's what we mean by the plus/minus sign in

3. ## Re: I need help-code and math

Originally Posted by grapes
The point is, what we do to the equation can affect the solution, and we sometimes have to account for those changes.

For instance, x=3 is a simple equation that has a simple solution. There is only one value for x that makes it true, and that value is 3. If we were to square both sides, we get which has two solutions, 3 and -3. By squaring both sides, we've introduced the possibility of a solution that wasn't a solution to the original. Similarly, when we take the square root of both sides, we have to remember to keep both solutions.

That's what we mean by the plus/minus sign in
That makes sense to me.

4. ## Re: I need help-code and math

Originally Posted by Eleftherios Karagiannis

I get that part, it's just like coding.
So we take the square roots of both sides, keeping both roots:

Now the parentheses are not necessary

And we can add 5 to both sides

Now we divide both sides by 3

So we end up with a lone x on the left hand side:

5. ## Re: I need help-code and math

Originally Posted by grapes

So we end up with a lone x on the left hand side:

I can follow all that, but highly doubt I could have gotten to x on my own. My goal is to get x all by itself, that helps me better understand, and maybe now I can.

So x can be or , right? Or am I getting ahead of my self?

6. ## Re: I need help-code and math

Originally Posted by Eleftherios Karagiannis
I can follow all that, but highly doubt I could have gotten to x on my own. My goal is to get x all by itself.
It takes a little practice. But you quickly see what has to be done to both sides of the equation so that the variable you are interested in ("x") gets maneuvered to one side by itself.
So x can be or , right?
Yes, exactly. And typically you should try them in the original equation to make sure:

Now, once you're given a formula like that, the steps to solving it are similar.

Another type of equation that we can solve is of the quadratic form, where each term is either a constant, has a factor of x, or has a factor of x squared:

7. ## Re: I need help-code and math

Originally Posted by grapes

I am going to try and solve this on my own, I like trying, even if I am wrong, I will learn along the way, hopefully.

I will let you know, either way.

8. ## Re: I need help-code and math

Originally Posted by grapes
It took decades, maybe millennia, but we learned that, in general, we should get all parts to one side. So we subtract one whole side from both sides:

which will leave zero on one side:

We can get rid of the parentheses, but notice we are subtracting each term:

Rearranging:

At this point, we could use the famous quadratic formula to solve this. That formula says that if , then the solution to it is

where, in our problem, a=9, b=-30, and c=16

But, that general solution was found by a method called completing the square, and I will solve our problem by that method.

OK so far?

9. ## Re: I need help-code and math

Originally Posted by grapes
To "complete the square" we add and divide a value to the equation. This value will allow a portion of the expression to be represented as an algebraic square

The value that we add and subtract is the second coefficient squared, divided by four times the first coefficient. In other words, , which is equal to 25:

The reason we do this is that makes the first three terms into a perfect algebraic square:

So we can replace those first three terms with the square:

Now we add 9 to both sides

And we've arrived at the same problem that we solved before:

10. ## Re: I need help-code and math

Wow, that was far harder than I could of imagined, it did not seem that hard when I first saw it. I could not get that far in trying, I failed, that is okay, it was only thinking. Now that I have seen the answer, I have a long road ahead of me to learn this.

I hope the coding part is easier than this, otherwise, I have two long and hard roads in front of me. All learning is like that, after all.

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