All you math junkies...

[Will Wood]

Solve for X. It's all variables. (If you might have confused b for 6 or whatever.)

I hate math (Kinda good at it though), but somehow I find my self trying to solve this now. Classmate came up to me with this problem.

Look at the problem below.. my image was way too big.

[LittMag]

will really quickly I get
x= sqroot((b^2r^2 + 2axc - c^2)/(b^2 - a^2))

[Collegeboy]

You can check my math I don't think I made a mistake but I am good at simple mistakes so please check.

[Collegeboy]

Here you go

[tobz]

umm

yeah, the square root of (a^2 + b^2) is not (a + b)

that would mean the square root of (3^2 + 6^2) = 9, when it actually is supposed to be the square root of (9+36) which = the square root of 45, which can further be simplified to 3*sqrt(5)

I will solve this for you, give me a bit of time.

T.J.

[Collegeboy]

Yes it is.

Just think about it.

Square root of a^2 is A. For it is saying the square of a times a, which is a.

[tobz]

right

It's like a recursive description. I think he wants to try and SOLVE for x. Which means he wants an explicit definition. I have one done. :) Using the good ol' quadratic formula.
.
...... this part means "plus or minus"
.................. v | this entire part is underneath a radical sign |
......................... __________________________________
....... -2ac +/- v (2ac)^2 - 4(b^2 - a^2)((-b^2)(r^2)-c^2)
.X = -----------------------------------------------------------------------
........................ 2(b^2 - a^2)
.
.
The dots on the left are used just to align everything, please disregard them. The dashed line in the center is a division line. And all of the carrot marks "^" are powers. Meaning b^2 = b-squared. Let me know if you have any problems. Thanks!
T.J.

[tobz]

this might be clearer.. lol

hopefully...

-T.J.

[DaosBeoulve]

Originally posted by Collegeboy
Yes it is.

Just think about it.

Square root of a^2 is A. For it is saying the square of a times a, which is a.

What tobz said.

If you square (a+b), you do not get (a^2+b^2), you get (a^2+2ab+b^2), simply because squaring (a+b) implies (a+b)(a+b).

Tobz, I have one question about your answer. Original problem said x=(that square root I don't feel like typing all out)

I agree with your quadratic answer, when you simplify out the right side, but did you factor in the left?

My brain refuses to work right now to work through all the algebra, and mathematica kicked me when I tried to get it to solve it for me=)

[tobz]

yup!

I started by squaring both sides and that is where the X^2 term comes from. You also get another X^2 term by multiplying out what is under the radical, which is how I ended up with "(a^2 + b^2)x^2" I just factored the X^2 out and then a^2 + b^2 ends up as the coeff. Which I used in the quadratic formula as "a". Thanks for asking, because you remined me, when I squared both sides, some incorrect or imaginary roots may come up :) Make sure you double check your solution

Thanks.
T.J.

[-=Squid=-]

Technically isnt it already solved since X is alone on one side? Im 99% its a trick question...

[tobz]

umm..

I just solved it. backup my work if you'd like, in fact I'd appreciate it. If I find a scanner I'll scan my paper in, but I'm NOT writing all my work in paint :) lol

X is solved for, but as you can see X is on the right side as well. Which means that it is implicity defined, but what I did was Explicitly Define it. That way x is NOT dependent upon x on the right side. It was only solved for using a slight recursive or implicit definition, and now I just solved it so it was explicit.

Thanks.
T.J.

[-=Squid=-]

Whoops, I didnt realise it was on both sides....my mistake, sorry about that...ignore my ignorance