5
$\begingroup$

Problem Statement:-

Solve the equation $$x^2+\dfrac{9x^2}{(x+3)^2}=27$$


I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value.

These are the things that I have tried to turn the given equation into a quadratic equation.

$$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{1}{\left(\dfrac{x}{3}+1\right)^2}=3\left(\dfrac{3}{x}\right)^2$$ $$\text{OR}$$ $$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{\left(\dfrac{3}{x}\right)^2}{\left(1+\dfrac{3}{x}\right)^2}=3\left(\dfrac{3}{x}\right)^2$$

$\endgroup$
  • $\begingroup$ I was trying to make such a substitution ot turn the given quartic to a quadratic not just by simple algebraic manipulation $\endgroup$ – user350331 Nov 18 '16 at 16:39
  • $\begingroup$ @J. M. isn't a mathematician - Am I not supposed to use \dfrac in question titles, please tell me the reason too please. I think that I was advised not to do so somewhere else too, but with some frivolous searching I couldn't find it. So can you tell me, what is the reason for the same. $\endgroup$ – user350331 Jan 28 '17 at 12:46
  • $\begingroup$ Stuff that explicitly or implicitly uses \displaystyle, \dfrac included, takes up more space that it normally would in the main page. In the interest of not being greedy with space, we avoid such constructs in titles. $\endgroup$ – J. M. is a poor mathematician Jan 28 '17 at 16:04
10
$\begingroup$

HINT:

Using $a^2+b^2=(a-b)^2+2ab,$ $$x^2+\left(\dfrac{3x}{x+3}\right)^2=\left(x-\dfrac{3x}{x+3}\right)^2+2\cdot x\cdot\dfrac{3x}{x+3}=\left(\dfrac{x^2}{x+3}\right)^2+6\cdot\dfrac{x^2}{x+3}$$

Generalization :

For $a^2+b^2=k$

If $\dfrac{ab}{a+b}=c$ where $c$ is a non-zero finite constant,

$$\implies k=(a+b)^2-2ab=(a+b)^2-2(a+b)c$$ $$\iff(a+b)^2-2(a+b)c-k=0$$ which isa Quadratic equation in $a+b$

Can you recognize $a,b$ here?

If $\dfrac{ab}{a-b}=c$ use $a^2+b^2=(a-b)^2+2ab$

$\endgroup$
  • $\begingroup$ What happens if $c$ isn't a constant? By the way, brilliant generalization! []+1 $\endgroup$ – Dr. Mathva Apr 9 at 19:23
3
$\begingroup$

Simplifying and reducing we get

$$x^4+6x^3-9x^2-162x-243=0\,\text{and}\,x\neq 3$$

Let's look for a factorisation in the form

$$x^4+6x^3-9x^2-162x-243=(x^2+ax+b)(x^2+cx+d)$$

Developing and comparing the coefficients we get

$$\begin{align}a+c=&6\\b+d+ac=&-9\\ad+bc=&-162\\bd=&-243\end{align}$$

We obviously look first for integer coefficients. We start with $-243=-3^5$ and we test the various combinations for $b,d$.

We get the following set that works

$$\begin{align}a=&-3\\b=&-9\\c=&9\\d=&27\end{align}$$

And now we're left with two quadratics

$$\begin{align}x^2-3x-9=&0\\x^2+9x+27=&0\end{align}$$

And the roots are

$$\begin{align}x_1=&3\cdot{1+\sqrt{5}\over 2}\\x_2=&3\cdot{1-\sqrt{5}\over 2}\\x_3=&3\cdot{-3+i\sqrt{3}\over2}\\x_4=&3\cdot{-3-i\sqrt{3}\over 2}\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.