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