Algebra Problem of The Day - 5

[BigBeachBanana]

[math]\large \sqrt{x-\frac{1}{x}} + \sqrt{1-\frac{1}{x}}=x[/math]Solve for [imath]\large x.[/imath]

 

[Harry_the_cat]

[math]\large \sqrt{x-\frac{1}{x}} + \sqrt{1-\frac{1}{x}}=x[/math]Solve for [imath]\large x.[/imath]

Just wondering ... Is one of the minus signs meant to be a plus?

 

[Steven G]

I can't do this one

 

[BigBeachBanana]

Just wondering ... Is one of the minus signs meant to be a plus?

I tripled-checked before I post...considering the number of mistakes I've been making. The OP is correct. It is a tougher one but can be solved algebraically. Here are some hints.

Spoiler: Hint#1

Conjugate

Spoiler: Hint#2

System of 2 equations

 

[Steven G]

I spent a good part of today working on this problem. The two hints helped

Spoiler

Clearly x>1

\(\displaystyle \sqrt{x-\dfrac{1}{x}}+ \sqrt{1-\dfrac{1}{x}} = x\)

Multiplying by conjugate yields \(\displaystyle x-1 = x(\sqrt{x-\dfrac{1}{x}}- \sqrt{1-\dfrac{1}{x}})\) or \(\displaystyle 1-\dfrac{1}{x}=\sqrt{x-\dfrac{1}{x}}- \sqrt{1-\dfrac{1}{x}}\)

Adding the 2 eqs yields \(\displaystyle x+1- \dfrac{1}{x} = 2\sqrt{x-\dfrac{1}{x}}\)

Let \(\displaystyle u = x-\dfrac{1}{x}\)

Then \(\displaystyle u+1 = 2\sqrt u\)
....
\(\displaystyle u=1\), a double root
\(\displaystyle x - \dfrac {1}{x} = 1\)
...
\(\displaystyle \dfrac{1+\sqrt 5}{2}\)

 

[Steven G]

@BigBeachBanana,
I will get you for making me spend hours on this one.
Steven

 

[Steven G]

You can't say that I didn't try. Besides in the end I got it! @Subhotosh Khan- do I get my long awaited raise now??
To the few who already knew how to do this problem thanks for not posting the solution.

 

[red12dog34]

For the existence of the square roots [math]x\in[-1,0)\cup[1,\infty)[/math]But the right hand member must be pozitive, so [math]x\in[1,\infty)[/math]The equation can be written as
[math]\sqrt{\frac{x^2-1}{x}}+\sqrt{\frac{x-1}{x}}=x\Leftrightarrow \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}\Leftrightarrow\\ \Leftrightarrow x^2+x-2+2\sqrt{x^3-x^2-x+1}=x^3\Leftrightarrow 2\sqrt{x^3-x^2-x+1}=\left(x^3-x^2-x+1\right)+1[/math]Let [math]\sqrt{x^3-x^2-x+1}=t[/math]Then [math]t^2-2t+1=0\Leftrightarrow (t-1)^2=0\Leftrightarrow t=1[/math]Then [math]\sqrt{x^3-x^2-x+1}=1\Leftrightarrow x^3-x^2-x=0\Leftrightarrow x^2-x-1=0[/math]The solutions are [math]x_{1.2}=\frac{1\pm\sqrt{5}}{2}[/math]But [math]x\in[1,\infty)[/math]so [math]x=\frac{1+\sqrt{5}}{2}[/math]

 

[BigBeachBanana]

Well done!

 

[BigBeachBanana]

@BigBeachBanana,
I will get you for making me spend hours on this one.
Steven

FYI: the problem got me for 2 days already.

 

[Steven G]

[math]\sqrt{\frac{x^2-1}{x}}+\sqrt{\frac{x-1}{x}}=x\Leftrightarrow \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}\Leftrightarrow\\ \Leftrightarrow x^2+x-2+2\sqrt{x^3-x^2-x+1}=x^3\Leftrightarrow 2\sqrt{x^3-x^2-x+1}=\left(x^3-x^2-x+1\right)+1[/math]

I actually had done exactly what you did but I failed to let t= \(\displaystyle \sqrt{x^3-x^2-x+1}\)
In the corner again!

 

[Steven G]

FYI: the problem got me for 2 days already.

Why would you do this to us mortals?? Some people, like red12dog34, have the ability to just write down the solution with ease but others.....