Find $x$, given: $x^2 + \frac{9x^2}{(x+3)^2} = 16$

Question

Here's an equation:

$$x^2 + \frac{9x^2}{(x+3)^2} = 16$$

First, I subtracted 16 from both sides and factored $x^2$ so I would get a quadratic equation, but with no success. Also, I can see that the equation can be rewritten as:

$$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$ But I can't see how can I use that information.

What should I do?

Answer 1

You can rewrite it as: \begin{align*} \left(\frac{x}{4}\right)^2 + \left(\frac{3x}{4(x+3)}\right)^2 & = 1. \end{align*} Let $\frac{x}{4}=\cos A$, then $\frac{3x}{4(x+3)}=\sin A$. Using $x=4 \cos A$ in the expression for $\sin A$, we get \begin{align*} 12 (\cos A -\sin A) & = 8 \sin 2A\\ 9(\cos A -\sin A)^2 & = 4 \sin ^22A\\ 9(1-\sin 2A)&=4 \sin^2 2A. \end{align*} Now you have a quadratic in $\sin 2A$. Can you proceed from here?

Comment: (added) This quadratic gives $\sin2A= -3$ or $\sin 2A=\frac{3}{4}$. The former cannot happen (for real values of $x$). The latter upon substituting back in terms of $x$ gives two real solutions $x=1\pm \sqrt{7}$.

Answer 2

Hint:

Multiply by $(x+3)^2$,

$$x^2(x+3)^2+9x^2=16(x+3)^2$$

or

$$x^4+6x^3+2x^2-96x-144=0.$$

Using a solver, the roots are irrational, but there is a rational factorization in two quadratic polynomials.

Answer 3

$$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$

multiply by $(x+3)^2$

$$(x+4)(x-4)(x+3)^2 + 9x^2 = 0$$ $$x^4+6x^3+2x^2-96x-144=0$$

$$(x^2-2x-6)(x^2+8x+24)=0$$

real solutions : $x=1\pm \sqrt{7}$

complex solutions : $x=-4\pm\iota 2\sqrt{2}$