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Find real solutions

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
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
Feb 7, 2012
2,707
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

Administrator
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Feb 24, 2012
13,775
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.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,685
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...:eek:

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!(Sun)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
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

Administrator
Staff member
Feb 24, 2012
13,775
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. (Yes)