# Find real solutions

#### anemone

##### MHB POTW Director
Staff member
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

##### MHB Oldtimer
Staff member
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.

#### MarkFL

Staff member
Let:

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

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.

#### anemone

##### MHB POTW Director
Staff member
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.
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...

Let:

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

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.
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!

#### anemone

##### MHB POTW Director
Staff member
Hey 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:

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

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:

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

is the only real root of the given equation.

#### MarkFL

Staff member
Hey 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:

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

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:

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

is the only real root of the given equation.
Quite clever, and alleviates the need for any root analyses.
The text passage contains a mathematical question and multiple solutions to that question. The question is to find all real solutions of a given equation, and the solutions provided by different users lead to the conclusion that the only real solution is \( a = 2 \).
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Find real solutions

#### anemone

##### MHB POTW Director
Staff member
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

##### MHB Oldtimer
Staff member
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.

#### MarkFL

Staff member
Let:

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

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.

#### anemone

##### MHB POTW Director
Staff member
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.
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...

Let:

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

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.
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!

#### anemone

##### MHB POTW Director
Staff member
Hey 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:

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

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:

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

is the only real root of the given equation.

#### MarkFL

Staff member
Hey 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:

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

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:

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

is the only real root of the given equation.
Quite clever, and alleviates the need for any root analyses.
user
The text passage contains a mathematical question and multiple solutions to that question. The question is to find all real solutions of a given equation, and the solutions provided by different users lead to the conclusion that the only real solution is \( a = 2 \).

TYPE: SINGLE_QUESTION_WITH_ANSWER