MHB Real Solutions for Equation: a-1

[anemone]

Find all real solutions of the equation $\sqrt{a-1}+\dfrac{\sqrt{a-1}}{a^2}+\dfrac{a^2}{a-1}=\dfrac{1}{\sqrt{a-1}}+\dfrac{a-1}{a^2}+\dfrac{a^2}{\sqrt{a-1}}$.

 

[Opalg]

[sp]Let $b = \sqrt{a-1}$. Then $b + \dfrac b{a^2} + \dfrac{a^2}{b^2} = \dfrac1b + \dfrac{b^2}{a^2} + \dfrac{a^2}b.$ Also, $b^2 = a-1$, $b>0$ (assuming that we take the square root to be the positive one) and $a>1$ (so that the square root is defined). Multiply through by $a^2b^2$ to get $$a^2b^3 + b^3 + a^4 = a^2b + b^4 + a^4b.$$ Write this as $$b(b^2-a^2) + a^2b(b^2-a^2) - (b^4 - a^4) = 0.$$ That factorises as $(b^2-a^2)(b + a^2b - a^2-b^2) = 0$, and then factorises further as $(b+a)(b-a)(a^2-b)(b-1) = 0.$ The first factor cannot be zero because $a$ and $b$ are both positive. If $b=a$ then the equation $b^2=a-1$ becomes $a^2-a+1=0$, which has no real solutions. Similarly if $b=a^2$ then we get the equation $a^4-a+1=0$, which also has no real solutions. Finally, if $b=1$ then $a=2$, and that is the only solution to the problem.[/sp]

 

[MarkFL]

Spoiler

Let:

$$0<u^2=a-1$$

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

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

which can be arranged and factored as:

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

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

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

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:

$$u\approx0.22602207985644908028$$

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:

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

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

$$a-1=1^2\implies a=2$$

is the only real root of the given equation.

 

[anemone]

[sp]Let $b = \sqrt{a-1}$. Then $b + \dfrac b{a^2} + \dfrac{a^2}{b^2} = \dfrac1b + \dfrac{b^2}{a^2} + \dfrac{a^2}b.$ Also, $b^2 = a-1$, $b>0$ (assuming that we take the square root to be the positive one) and $a>1$ (so that the square root is defined). Multiply through by $a^2b^2$ to get $$a^2b^3 + b^3 + a^4 = a^2b + b^4 + a^4b.$$ Write this as $$b(b^2-a^2) + a^2b(b^2-a^2) - (b^4 - a^4) = 0.$$ That factorises as $(b^2-a^2)(b + a^2b - a^2-b^2) = 0$, and then factorises further as $(b+a)(b-a)(a^2-b)(b-1) = 0.$ The first factor cannot be zero because $a$ and $b$ are both positive. If $b=a$ then the equation $b^2=a-1$ becomes $a^2-a+1=0$, which has no real solutions. Similarly if $b=a^2$ then we get the equation $a^4-a+1=0$, which also has no real solutions. Finally, if $b=1$ then $a=2$, and that is the only solution to the problem.[/sp]

Thank you Opalg for participating! I initially thought to share the solution of other that I considered to be a brilliant way to attack this problem but now I see there is no need to do so...

Spoiler

Let:

$$0<u^2=a-1$$

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

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

which can be arranged and factored as:

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

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

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

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:

$$u\approx0.22602207985644908028$$

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:

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

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

$$a-1=1^2\implies a=2$$

is the only real root of the given equation.

Hey MarkFL, my dearest and sweetest admin, I like your method so much as well! I think this is another great method to crack this problem! Thanks for your solution!(Sun)

 

[anemone]

Hey MarkFL,:) I see there is another way to prove the second factor in $$(u-1)\left(u^6+2u^4+2u^2-u+1 \right)$$ so that it is always greater than zero:

Spoiler

$$(u-1)\left(u^6+2u^4+2u^2-u+1 \right)=0$$

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:

$$a-1=1^2\implies a=2$$

is the only real root of the given equation.

 

[MarkFL]

Hey MarkFL,:) I see there is another way to prove the second factor in $$(u-1)\left(u^6+2u^4+2u^2-u+1 \right)$$ so that it is always greater than zero:

Spoiler

$$(u-1)\left(u^6+2u^4+2u^2-u+1 \right)=0$$

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:

$$a-1=1^2\implies a=2$$

is the only real root of the given equation.

Quite clever, and alleviates the need for any root analyses. (Yes)