x^2 +(x/(x+1))^2=3.All possible solution of x

1. Equation given

2. Clear the denominator

Multiplying every term by $(x+1)^2$ removes the fraction:

which becomes

3. Expand carefully

Remember $(x+1)^2 = x^2 + 2x + 1$.

• Left term: $x^2(x^2 + 2x +1) = x^4 + 2x^3 + x^2$.
• Add the extra $+x^2$ to get $x^4 + 2x^3 + 2x^2$.
• Right term: $3(x^2 + 2x +1) = 3x^2 + 6x + 3$.

4. Gather everything on one side

Now we have a quartic (degree-4) polynomial.
Direct factoring looks messy, so most students test simple values or use RRT (Rational Root Test) to guess integer roots. None satisfy; hence numeric/clever inspection is helpful.

5. Spotting a pattern: the golden ratio $\varphi$

Notice that $\varphi = \dfrac{1+\sqrt5}{2} \approx 1.618$ has the famous property $\varphi^2 = \varphi + 1$.
Plugging $x=\varphi$:

Adding: $2.618 + 0.382 = 3$ ✔️.
By symmetry, $x=-\dfrac{1}{\varphi}=\dfrac{1-\sqrt5}{2}\approx -0.618$ also works (check the sign: the fraction stays finite because $-0.618+1>0$).

6. Domain check

Remember $x \neq -1$, and neither solution equals $-1$. So both are valid.

🧒🏻 Imagine two squares

1. We draw a big square whose side is x. Its area is $x^2$.
2. Next, we draw a second square whose side is the funny fraction $\dfrac{x}{x+1}$. Its area is therefore $\left(\dfrac{x}{x+1}\right)^2$.

The puzzle says: When the areas of those two squares are added, the total must be 3.
So we want to find all numbers x that can make that happen.

But we must be careful: whenever we see a fraction with $(x+1)$ in the bottom, we cannot let $x=-1$, because division by zero is not allowed.

Step-by-Step Solution

1. Start with the equation

2. Clear denominators Multiply both sides by $(x+1)^2$:

3. Expand

4. Bring all terms to LHS

5. Observe the golden-ratio factor
   Notice $x=\varphi=\dfrac{1+\sqrt5}{2}$ makes $(\*)$ zero (verified in the explanation). Therefore $(x-\varphi)$ is a factor. Likewise, $x= -\dfrac{1}{\varphi}=\dfrac{1-\sqrt5}{2}$ is another root, so $(x + \dfrac{1}{\varphi})$ is also a factor.

6. Factor out the two real roots (long division/ synthetic division reveals the remaining quadratic has no further real roots). Finally, we keep only the real solutions.

Final Answer