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I am trying to solve the following equation, but I have no idea where to start. Can somebody point me in the right direction? ...well, if somebody knows how to solve it that would be great, otherwise hints would be great.$$\ h=\left(\dfrac{a+bx^2}{cx}\right)\left(\dfrac{d+fx^2}{gx}\right)\left(\left(\dfrac{a+bx^2}{cx}\right)\left(\dfrac{d+fx^2}{gx}\right)+i\right)-\left(\dfrac{a+bx^2}{cx}\right)^2-\left(\dfrac{d+fx^2}{gx}\right)^2 $$

I need to solve for x, where all other variables are known. I would also like to point out I can use a computer to help if that is necessary.

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Do you need an analytic solution, or a numeric value is enough? –  yohBS Jan 22 '12 at 18:27
I would love a sep -by-step solution in order to learn, but an x=... solution would be more than enough. –  Adrien Hingert Jan 22 '12 at 18:31
Lord Wolfram is angry with me! –  user21436 Jan 22 '12 at 18:41
@KannappanSampath, it seems Wolfram doesn't know how to interpret the question :( –  Adrien Hingert Jan 22 '12 at 18:54
I up-voted this question late (very late, lol) because I feel it does qualify as very appropriate for the question format of this site. I had forgot to upvote it beforehand, lol. –  000 Jan 23 '12 at 20:50
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2 Answers

up vote 5 down vote accepted

Let there be the following substitutions: $$\begin{align} &w=x^2 &\\ &u=a+bw &j=d+fw \end{align}$$ The equation reduces to: $$\begin{align} h&=\frac{u}{cx}\frac{j}{gx}\left(\frac{u}{cx}\frac{j}{gx}+i\right)-\left(\frac{u}{cx}\right)^2-\left(\frac{j}{gx}\right)^2\\ &=\left(\frac{uj}{cgx^2}\right)^2+\frac{uj}{cgx^2}i-\left(\frac{u}{cx}\right)^2-\left(\frac{j}{gx}\right)^2 \end{align}$$ Because of my slight dislike of fractional powers (I did not sub in $w^{\frac{1}{2}}$ for $x$), we can recall what $w$ is and restate the equation: $$h=\frac{u^2j^2}{c^2g^2w^2}+\frac{uj}{cgw}i-\frac{u^2}{c^2w}-\frac{j^2}{g^2w}$$ Getting a polynomial in $w$ seems to be the most enlightening exercise, so multiply through by $w^2$: $$hw^2=\frac{u^2j^2}{c^2g^2}+\frac{uj}{cg}iw-\frac{u^2}{c^2}w-\frac{j^2}{g^2}w$$ Rearranging this gives us a cleaner expression: $$-hw^2-\left(\frac{u^2}{c^2}+\frac{j^2}{g^2}\right)w+\frac{uj}{cg}iw+\frac{u^2j^2}{c^2g^2}=0$$ Things will be slightly simpler if we multiply through by $c^2g^2$: $$-hc^2g^2w^2-(u^2g^2+j^2c^2)w+ujcgwi+u^2j^2=0$$ Since $uj$, $u^2$, $j^2$ and $u^2j^2$ are the important things in the equation, let's simplify them: $$\begin{align} uj&=(a+bw)(d+fw)\\ &=bfw^2+(af+bd)w+ad\\ u^2&=(a+bw)^2\\ &=b^2w^2+2abw+a^2\\ j^2&=(d+fw)^2\\ &=f^2w^2+2dfw+d^2\\ u^2j^2&=(b^2w^2+2abw+a^2)(f^2w^2+2dfw+d^2)\\ &=(b^2f^2)w^4+(2abf^2+2b^2df)w^3+(a^2f^2+b^2d^2+4abdf)w^2+(2a^2df+2abd^2)w+(a^2d^2) \end{align} $$ It may help to work $u^2j^2$ with the following substitutions in mind: $$\begin{align} &\alpha_{2}=b^2 &\alpha_{1}=2ab &\alpha_{0}=a^2\\ &\beta_{2}=f^2 &\beta_{1}=2df &\alpha_{0}=d^2 \end{align}$$ The equivalent equation is, thusly: $$u^2j^2=\alpha_2 \beta_2 w^4+(\alpha_1 \beta_{2}+\alpha_{2} \beta_{1})w^3+(\alpha_0 \beta_2+\alpha_1 \beta_1+\alpha_2 \beta_0)w^2+(\alpha_0 \beta_1+\alpha_1 \beta_0)w+(\alpha_0 \beta_0)$$ It is now a task of digging through this algebra to separate the terms. . . Firstly, look at the 2nd coefficient: $$\begin{align} u^2g^2+j^2c^2&=(b^2w^2+2abw+a^2)g^2+(f^2w^2+2dfw+d^2)c^2\\ &=(b^2g^2+f^2c^2)w^2+(2abg^2+2dfc^2)w+(a^2g^2+d^2c^2) \end{align}$$ Next, look at the third term: $$\begin{align} ujcgwi&=[bfw^2+(af+bd)w+ad]cgwi\\ &=(bfcgi)w^3+(afcgi+bdcgi)w^2+(adcgi)w \end{align}$$ Now, combining all of these forms is the nearly final step (Let the equation be abbreviated with an E): $$ \begin{align} E&=-hc^2g^2w^2-((b^2g^2+f^2c^2)w^2+(2abg^2+2dfc^2)w+(a^2g^2+d^2c^2))w\\ &+(bfcgi)w^3+(afcgi+bdcgi)w^2+(adcgi)w\\ &+(b^2f^2)w^4+(2abf^2+2b^2df)w^3\\ &+(a^2f^2+b^2d^2+4abdf)w^2+(2a^2df+2abd^2)w+(a^2d^2)\\ &=(b^2f^2)w^4\\ &+(2abf^2+2b^2df)w^3-(b^2g^2+f^2c^2)w^3++(bfcgi)w^3\\ &+(a^2f^2+b^2d^2+4abdf)w^2-(2abg^2+2dfc^2)w^2+(afcgi+bdcgi)w^2-(hc^2g^2)w^2\\ &-(a^2g^2+d^2c^2)w+(2a^2df+2abd^2)w+(adcgi)w\\ &+a^2d^2 \end{align} $$ This shows us that the coefficients of the terms are as follows: $$ \begin{align} C_4&=b^2f^2\\ C_3&=2abf^2+2b^2df-b^2g^2-f^2c^2+bfcgi\\ C_2&=a^2f^2+b^2d^2+4abdf-2abg^2-2dfc^2+afcgi+bdcgi-hc^2g^2\\ C_1&=-a^2g^2-d^2c^2+2a^2df+2abd^2+adcgi\\ C_0&=a^2d^2 \end{align} $$

Thus, we have the following quartic in $w$: $$C_4w^4+C_3w^3+C_2w^2+C_1w+C_0=0$$

I leave the rest as an exercise to the reader. . .

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Thanks, I'll let you know if this is too complicated for me :) –  Adrien Hingert Jan 23 '12 at 17:31
@AdrienHingert, I recommend checking my equations, by the way. My process is correct--it's just that I may have flubbed a symbol here or there. :) BTW, you can extend this problem to solving a quartic (generally) or you have the option of going straight to the quartic formula. Understanding how to derive the general solution of quartics (and cubics) is a bit difficult, though, IMHO. It's still a very fun procedure that enlightens you on how cool some basic algebraic techniques are. –  000 Jan 23 '12 at 20:48
I'm looking at Wikipedia and Wolfram aplpha to understand how to solve a quartic although I might need to get back to you on this one :) Thanks for the help up to now, this was great! –  Adrien Hingert Jan 24 '12 at 17:09
@AdrienHingert, I'm really happy that I could help you. My abilities are limited and it brings me great joy to be able to make a substantial contribution to math.stackexchange. :) Goodluck in future mathematical endeavors! –  000 Jan 24 '12 at 21:49
I managed to finally go through it all. There was one small mistake - you left out bfcgi out of the C3 list coefficients -, but I cannot thank you enough for having helped me so much! –  Adrien Hingert Jan 26 '12 at 13:22
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Multiply both sides by $(cgx^2)^2$,

$h (c g x^2)^2 = (a+bx^2)(d+fx^2)\left( (a+bx^2)(d+fx^2)+i (c g x^2)\right)-(a+bx^2)^2(gx)^2-(d+fx^2)^2(cx)^2$

One can see this is merely a quartic equation in $x^2$. There is an analytic expression for x using the quartic formula, but it would be very messy.

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Thanks, I'll go through it, but I'm not sure I would know how to proceed from here :) –  Adrien Hingert Jan 23 '12 at 17:30
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