- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,756

Solve \(\displaystyle x^2+\frac{x^2}{(x+1)^2}=3\)

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,756

Solve \(\displaystyle x^2+\frac{x^2}{(x+1)^2}=3\)

- Moderator
- #2

- Feb 7, 2012

- 2,725

Multiply by $(x+1)^2$ to get $(x^2-3)(x+1)^2 + x^2 = 0$, or $x^4+2x^3-x^2-6x-3=0.$ A graphing calculator shows that the quartic equation has two real roots, approximately $1.62$ and $-0.62.$ The only hope of finding the exact values is if the quartic has two quadratic factors. Since the sum of those two roots looks as though it is close to $1$, and their product close to $-1$, one of the factors ought to be $x^2 -x - 1.$ Sure enough, you find that $x^4+2x^3-x^2-6x-3=(x^2 -x - 1)(x^2+3x+3).$ So the (real) roots are those of the first factor, namely $x = \frac12\bigl(1\pm\sqrt5\bigr).$Solve \(\displaystyle x^2+\frac{x^2}{(x+1)^2}=3\)

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,756

Thanks Opalg, for your analysis of the problem and the correct solutions as well.

I feel somewhat reticent to post my 'silly' method but at the same time, I wanted to know if this is what we could actually do to solve an equation...

My solution:

Let \(\displaystyle k=x+1\), we now have

\(\displaystyle (k-1)^2+\frac {(k-1)^2}{k^2}=3\)

\(\displaystyle k^2(k-1)^2+(k-1)^2=3k^2\)

\(\displaystyle (k-1)^2(k^2+1)=3k^2\)

We see that one of the possibilities that this equation can be rewritten is in the following manner:

\(\displaystyle (k-1)^2=k\) and \(\displaystyle (k^2+1)=3k\)

For the first case, we simplify it to get:

\(\displaystyle k^2-3k+1=0\)

And for the second case, we also do the simplification and obtain:

\(\displaystyle k^2-3k+1=0\)

This means if we solve the equation \(\displaystyle k^2-3k+1=0\), we will also solve the original equation.

Thus, \(\displaystyle k=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}\)

Hence:

\(\displaystyle x=\frac{3\pm \sqrt{5}}{2}-1=\frac{1\pm \sqrt{5}}{2}\)

Edit: I have actually looked at all of the factor pair combinations in order to rewrite the equation \(\displaystyle (k-1)^2(k^2+1)=3k^2\) as two equivalent equations, and the one that I gave above was the only combination yielding real solutions.

I feel somewhat reticent to post my 'silly' method but at the same time, I wanted to know if this is what we could actually do to solve an equation...

My solution:

Let \(\displaystyle k=x+1\), we now have

\(\displaystyle (k-1)^2+\frac {(k-1)^2}{k^2}=3\)

\(\displaystyle k^2(k-1)^2+(k-1)^2=3k^2\)

\(\displaystyle (k-1)^2(k^2+1)=3k^2\)

We see that one of the possibilities that this equation can be rewritten is in the following manner:

\(\displaystyle (k-1)^2=k\) and \(\displaystyle (k^2+1)=3k\)

For the first case, we simplify it to get:

\(\displaystyle k^2-3k+1=0\)

And for the second case, we also do the simplification and obtain:

\(\displaystyle k^2-3k+1=0\)

This means if we solve the equation \(\displaystyle k^2-3k+1=0\), we will also solve the original equation.

Thus, \(\displaystyle k=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}\)

Hence:

\(\displaystyle x=\frac{3\pm \sqrt{5}}{2}-1=\frac{1\pm \sqrt{5}}{2}\)

Edit: I have actually looked at all of the factor pair combinations in order to rewrite the equation \(\displaystyle (k-1)^2(k^2+1)=3k^2\) as two equivalent equations, and the one that I gave above was the only combination yielding real solutions.

Last edited:

- Jan 25, 2013

- 1,225

proceed as Opalg's solution : factoring f(x)Multiply by $(x+1)^2$ to get $(x^2-3)(x+1)^2 + x^2 = 0$, or $x^4+2x^3-x^2-6x-3=0.$ A graphing calculator shows that the quartic equation has two real roots, approximately $1.62$ and $-0.62.$ The only hope of finding the exact values is if the quartic has two quadratic factors. Since the sum of those two roots looks as though it is close to $1$, and their product close to $-1$, one of the factors ought to be $x^2 -x - 1.$ Sure enough, you find that $x^4+2x^3-x^2-6x-3=(x^2 -x - 1)(x^2+3x+3).$ So the (real) roots are those of the first factor, namely $x = \frac12\bigl(1\pm\sqrt5\bigr).$

$f(x)=x^4+2x^3-x^2-6x-3=x^4+2x^3-3x-x^2-3x-3$

$=x(x^3+2x^2-3)-(x^2+3x+3)$

$=x(x^2+3x+3)(x-1)-(x^2+3x+3)$

$=(x^2-x-1)(x^2+3x+3)=0$

So the (real) roots are those of the first factor, namely $x = \frac12\bigl(1\pm\sqrt5\bigr).$