- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,815

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,815

- Moderator
- #2

- Feb 7, 2012

- 2,740

- Admin
- #3

\(\displaystyle 0<u^2=a-1\)

After making the substitution and multiplying the result by \(\displaystyle u^2\left(u^2+1 \right)^2\) we obtain:

\(\displaystyle u^3\left(u^2+1 \right)^2+u^3+\left(u^2+1 \right)^4=u\left(u^2+1 \right)^2+u^4+u\left(u^2+1 \right)^4\)

which can be arranged and factored as:

\(\displaystyle (u-1)\left(u^6+2u^4+2u^2-u+1 \right)\)

Regarding the sextic factor (let's call if $f$), let's look at its derivative:

\(\displaystyle f'(u)=6u^5+8u^3+4u-1\)

Descartes' Rule of Signs tells us there is at most 1 positive real root. We should observe that it will be close to zero. Use of a numeric root finding technique puts this root at about:

\(\displaystyle u\approx0.22602207985644908028\)

We can see that the second derivative will be positive for all real $u$, hence this critical value is at a global minimum. We then find:

\(\displaystyle f_{\min}\approx f(0.22602207985644908028)\approx0.881502759171198736472582>0\)

Hence, we may conclude $f$ has no real roots, and so $u=1$ is the only root, which means:

\(\displaystyle a-1=1^2\implies a=2\)

is the only real root of the given equation.

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,815

Thank you

Hey

\(\displaystyle 0<u^2=a-1\)

After making the substitution and multiplying the result by \(\displaystyle u^2\left(u^2+1 \right)^2\) we obtain:

\(\displaystyle u^3\left(u^2+1 \right)^2+u^3+\left(u^2+1 \right)^4=u\left(u^2+1 \right)^2+u^4+u\left(u^2+1 \right)^4\)

which can be arranged and factored as:

\(\displaystyle (u-1)\left(u^6+2u^4+2u^2-u+1 \right)\)

Regarding the sextic factor (let's call if $f$), let's look at its derivative:

\(\displaystyle f'(u)=6u^5+8u^3+4u-1\)

Descartes' Rule of Signs tells us there is at most 1 positive real root. We should observe that it will be close to zero. Use of a numeric root finding technique puts this root at about:

\(\displaystyle u\approx0.22602207985644908028\)

We can see that the second derivative will be positive for all real $u$, hence this critical value is at a global minimum. We then find:

\(\displaystyle f_{\min}\approx f(0.22602207985644908028)\approx0.881502759171198736472582>0\)

Hence, we may conclude $f$ has no real roots, and so $u=1$ is the only root, which means:

\(\displaystyle a-1=1^2\implies a=2\)

is the only real root of the given equation.

- Thread starter
- Admin
- #5

- Feb 14, 2012

- 3,815

Regarding the sextic factor (let's call if $f$),

$\begin{align*}u^6+2u^4+2u^2-u+1&=(u^6+2u^4+u^2)+(u^2-u+1)\\&=u^2(u^4+2u^2+1)+\left(u-\dfrac{1}{2} \right)^2+1-\dfrac{1}{4}\\&=u^2(u^2+1)^2+\left(u-\dfrac{1}{2} \right)^2+\dfrac{3}{4}\\& >0 \end{align*}$

Hence, we may conclude $f$ has no real roots, and so $u=1$ is the only root, which means:

\(\displaystyle a-1=1^2\implies a=2\)

is the only real root of the given equation.

- Admin
- #6

Quite clever, and alleviates the need for any root analyses.MarkFL, I see there is another way to prove the second factor in \(\displaystyle (u-1)\left(u^6+2u^4+2u^2-u+1 \right)\) so that it is always greater than zero:

Regarding the sextic factor (let's call if $f$),

$\begin{align*}u^6+2u^4+2u^2-u+1&=(u^6+2u^4+u^2)+(u^2-u+1)\\&=u^2(u^4+2u^2+1)+\left(u-\dfrac{1}{2} \right)^2+1-\dfrac{1}{4}\\&=u^2(u^2+1)^2+\left(u-\dfrac{1}{2} \right)^2+\dfrac{3}{4}\\& >0 \end{align*}$

Hence, we may conclude $f$ has no real roots, and so $u=1$ is the only root, which means:

\(\displaystyle a-1=1^2\implies a=2\)

is the only real root of the given equation.