- Thread starter
- Admin
- #1
- Feb 14, 2012
- 3,815
Find all real solutions of the equation $\sqrt{a-1}+\dfrac{\sqrt{a-1}}{a^2}+\dfrac{a^2}{a-1}=\dfrac{1}{\sqrt{a-1}}+\dfrac{a-1}{a^2}+\dfrac{a^2}{\sqrt{a-1}}$.
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 $b = \sqrt{a-1}$. 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.
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!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.
Quite clever, and alleviates the need for any root analyses.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.