Please could someone help me with this?

Is it possible to rearrange the following equation so that it is expressed in terms of x? Have been trying for ages without luck.

x^2 = b^2 + (c - (ac/(a+x)))^2

Any help would be much appreciated.

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- Jan 21st 2011, 03:05 PMEdCoventryRearrange in terms of x?
Please could someone help me with this?

Is it possible to rearrange the following equation so that it is expressed in terms of x? Have been trying for ages without luck.

x^2 = b^2 + (c - (ac/(a+x)))^2

Any help would be much appreciated. - Jan 21st 2011, 03:08 PMEdCoventry
Apologies if I have posted this in the wrong place.

- Jan 21st 2011, 03:19 PMe^(i*pi)
$\displaystyle x^2 = b^2 + \left(c - \dfrac{ac}{a+x}\right)^2$

A good place to start would be to use foil to expand that squared term

$\displaystyle \left(c - \dfrac{ac}{a+x}\right)^2 = c^2 - \dfrac{2ac^2}{a+x} + \dfrac{a^2c^2}{(a+x)^2}$

You should then be able to multiply through by $\displaystyle (a+x)^2$ to clear the fraction

$\displaystyle x^2(a+x)^2 = b^2(a+x)^2 + c^2(a+x)^2 - 2ac^2(a+x) + a^2c^2$

You will eventually have to expand it and it won't be pretty but it should be simple enough using FOIL and the distributive property - Jan 21st 2011, 03:24 PMEdCoventry
Thanks for that. I'll have a go but think I've bitten off more than I can chew with this one. My mathematical capabilites are not great.

- Jan 21st 2011, 03:25 PMtonio

$\displaystyle \displaystyle{x^2=b^2+\left(c-\frac{ac}{a+x}\right)^2\Longrightarrow x^2-\left(c-\frac{ac}{a+x}\right)^2=b^2\Longrightarrow}$

$\displaystyle \displaystyle{\Longrightarrow \left(x-\left(c-\frac{ac}{a+x}\right)\right)\left(x+\left(c-\frac{ac}{a+x}\right)\right)=b^2\Longrightarrow}$

$\displaystyle \displaystyle{\Longrightarrow\left(x-\frac{cx}{a+x}\right)\left(x+\frac{cx}{a+x}\right) =b^2\Longrightarrow \left(x^2+(a-c)x\right)\left(x^2+(a+c)x\right)=b^2x+ab^2}$

$\displaystyle \displaystyle{\Longrightarrow x^4+2ax^3+\left(a^2-b^2-c^2)x^2+ab^2=0}$ ,

and you get a rather nasty quartic equation which nevertheless can be solved with the help

of Ferrari-Cardano formulae (pretty frightening creatures!), or perhaps you know some relation

between the coefficients $\displaystyle a,b,c$ which could simplify the above (for example, $\displaystyle a=0$ could make things waaaaaay simpler)

Tonio - Jan 21st 2011, 03:45 PMEdCoventry
Attachment 20540Attachment 20540No, definitely out of my depth. May have to try and find a simpler method. Was trying to find an equation to determine x on the attached diagram given a,b and c. Came up with the above. Is there a simpler approach to this?

Thanks again. - Jan 21st 2011, 03:53 PMtonio
- Jan 21st 2011, 04:02 PMEdCoventry
Tonio,

Oops, sorry about that. New to this. Should it be a in (a + b)^2 though? I thought that would be different term?

Apologies, please ignore, just seen your reply. - Jan 21st 2011, 04:07 PMtonio

You sent the sketch of a straight-angle triangle with hypotenuse equal to $\displaystyle a+x$ , vertical leg equal to $\displaystyle c$ and

horizontal leg equal to $\displaystyle b+?$ . It's the "?" part that we don't know what prevents from us to use Pythagoras Theorem

Tonio - Jan 21st 2011, 04:11 PMEdCoventry
Thanks Tonio. Looks like there's no simple solution then. Am left with trying to solve the nasty quartic equation... I think beyond the limit of my abilities

- Jan 22nd 2011, 02:33 AMEdCoventry
- Jan 22nd 2011, 03:11 AMe^(i*pi)
Yes it should. I made an expansion error (which is why you don't try complex problems after midnight)

$\displaystyle x^2(a+x)^2 = b^2(a+x)^2 + c^2(a+x)^2 - 2ac^2(a+x) + a^2c^2$

$\displaystyle x^2(a+x)^2 = x^2(a^2+2ax+x^2) = a^2x^2+2ax^3+x^4$

$\displaystyle b^2(a+x)^2 = b^2(a^2+2ax+x^2) = a^2b^2+2ab^2x+x^2b^2$

$\displaystyle c^2(a+x)^2 = c^2(a^2+2ax+x^2) = a^2c^2+2ac^2x+c^2x^2$

$\displaystyle 2ac^2(a+x) = 2a^2c^2+2ac^2x$

-----------------------------

$\displaystyle a^2x^2+2ax^3+x^4 = (a^2b^2+2ab^2x+x^2b^2)+ (a^2c^2+2ac^2x+c^2x^2) - (2a^2c^2+2ac^2x) + a^2c^2$

Collect those with an x term on the right hand side.

$\displaystyle [(2ab^2x +x^2b^2) + (2ac^2x+c^2x^2) - 2ac^2x] + (a^2b^2+a^2c^2-2a^2c^2+a^2c^2)$

Cancelling gives

$\displaystyle 2ab^2x +x^2b^2 + c^2x^2 + a^2b^2$

This is equal to the LHS

$\displaystyle a^2x^2+2ax^3+x^4 = 2ab^2x +x^2b^2 + c^2x^2 + a^2b^2$

$\displaystyle x^4+2ax^3+a^2x^2-2ab^2x-x^2b^2-c^2x^2 - a^2b^2 = 0$

Collect like terms

$\displaystyle x^4 + 2ax^3 + (a^2-b^2-c^2)x^2 - 2ab^2x - a^2b^2 = 0$

Which is a nasty quartic equation but will probably be easier if you use an iterative method - Jan 22nd 2011, 03:28 AMEdCoventry
Excellent, thanks for that. When I expand that I get:

x^4 + 2ax^3 + (a^2 - b^2 - c^2)x^2 - (2ab^2)x - (a^2)(b^2) = 0

Is this right as it's different to Tonio's expansion (also apologies for the format, haven't worked that out yet).

Thanks again. - Jan 22nd 2011, 03:35 AMe^(i*pi)
I edited working into my above post, it

*looks*right but it would be only too easy to make a mistake but at least we get the same answer! - Jan 22nd 2011, 03:49 AMEdCoventry
Brilliant. That's what I got. The only difference being the + 2ab^2x in the last step. Should that be - 2ab^2x.