# Solve an equation of 4th degree

Solve the following equation: $$x^2+\dfrac{81x^2}{(9+x)^2}=40.$$

Unfortunately I have no ideas because after expanding I get an equation of 4 degree.

• You should include the expansion (to degree $4$) in your Question and indicate what about the expanded polynomial equation gave you difficulty. – hardmath Apr 19 '16 at 12:22
• Every exponent is even. $t=x^2$ should help a lot. – Antitheos Apr 19 '16 at 12:36
• @Antitheos, $(9+x)^2=81+18x+x^2$ has an odd exponent. – Barry Cipra Apr 19 '16 at 12:40
• Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Apr 19 '16 at 16:58
• Just FYI. Fully expanded: $$x^4+18x^3+122x^2-720x-3240=0.$$ – stackErr Apr 19 '16 at 20:12

$$x^2+\dfrac{81x^2}{(9+x)^2}=40$$. $$x^2-\frac{18x^2}{x+9}+\frac{81x^2}{(x+9)^2}+\frac{18x^2}{x+9}=40$$ $$\left( x-\frac{9x}{x+9}\right)^2+\frac{18x^2}{x+9}=40$$ $$\left( \frac{x^2}{x+9}\right)^2+\frac{18x^2}{x+9}=40$$ Let $$\frac{x^2}{x+9}=t$$ Then $t^2+18t-40=0$. Then $t=-20$ or $t=2$.

$\frac{x^2}{x+9}=-20$ or $\frac{x^2}{x+9}=2$

$x^2+20x+180=0$ or $x^2-2x-18=0$

$x=1\pm\sqrt{19}$

Solve the following equation $A^2(x)+B^2(x)=c$, where $A(x)-B(x)=A(x)B(x)$ Then $$A^2(x)-2A(x)B(x)+B^2(x)+2A(x)B(x)=C$$ $$(A(x)-B(x))^2+2A(x)B(x)=c$$ Then $A(x)-B(x)=A(x)B(x)=t$

For example:

• It'd be interesting to add how you come up with the process. Once one guesses the correct substitution, it becomes trivial, but the hard part is to figure it out. – Petr Pudlák Apr 19 '16 at 11:57
• @petr-pudl%c3%a1k: it seems to me that in this particular case one is expected to see that both leftmost items are square of something so one is expected to insert +- of 2*sqrt(a)*sqrt(b) - to seeif it does something interesting, not that it necessarily leads to solution. After all, solving an equation analytically is improbable in common case. – Euri Pinhollow Apr 19 '16 at 17:48

Here is another way to solve it.

Fully expanding it gives: $$x^2(9+x)^2 + 81x^2 = 40(9+x)^2$$ $$(x^2-40)(9+x)^2 + 81x^2 = 0$$ $$(x^2-40)(81+18x+x^2)+81x^2 = 0$$ $$x^4+18x^3+122x^2-720x-3240=0$$

Important step: $$(x^2-2x-18)(x^2+20x+180)=0$$

The last step is tricky but if you assume that a factorization of the form exists: $$(ax^2 + bx + c)(dx^2 + ex + f) = 0$$

then mapping the coeffeicients to the corresponding powers of x gives: $$(ad)x^4 + (ae + db)x^3 + (af + dc + be)x^2 + (bf + ec)x + fc = 0$$

You have the following equations to solve:

$$(a*d) = 1$$ a=d=1 (or -1) for simplicity

So we have 4 equations and 4 unknowns $$(e+b) = 18$$ $$(f + c + be) = 122$$ $$(b*f + e*c) = -720$$ $$(f*c) = -3240$$

These are tough and long to solve by hand but eventually you get:

$$a = 1, b = -2, c = -18, d = 1, e = 20, f = 180$$

• The system of equations whose solution you haven't show also lead to 4th degree equation or a cubic. – Shubhashish Jan 7 '17 at 15:16