Algebra Problem of The Day - 5

BigBeachBanana

BigBeachBanana

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x−1x+1−1x=x\large \sqrt{x-\frac{1}{x}} + \sqrt{1-\frac{1}{x}}=xSolve for x.\large x.
 
Harry_the_cat said:
Just wondering ... Is one of the minus signs meant to be a plus?
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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.

Conjugate
System of 2 equations
 
I spent a good part of today working on this problem. The two hints helped

Clearly x>1

x−1x+1−1x=x\displaystyle \sqrt{x-\dfrac{1}{x}}+ \sqrt{1-\dfrac{1}{x}} = x

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

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

Let u=x−1x\displaystyle u = x-\dfrac{1}{x}

Then u+1=2u\displaystyle u+1 = 2\sqrt u
....
u=1\displaystyle u=1, a double root
x−1x=1\displaystyle x - \dfrac {1}{x} = 1
...
1+52\displaystyle \dfrac{1+\sqrt 5}{2}
 
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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.
 
For the existence of the square roots x∈[−1,0)∪[1,∞)x\in[-1,0)\cup[1,\infty)But the right hand member must be pozitive, so x∈[1,∞)x\in[1,\infty)The equation can be written as
x2−1x+x−1x=x⇔x2−1+x−1=xx⇔⇔x2+x−2+2x3−x2−x+1=x3⇔2x3−x2−x+1=(x3−x2−x+1)+1\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)+1Let x3−x2−x+1=t\sqrt{x^3-x^2-x+1}=tThen t2−2t+1=0⇔(t−1)2=0⇔t=1t^2-2t+1=0\Leftrightarrow (t-1)^2=0\Leftrightarrow t=1Then x3−x2−x+1=1⇔x3−x2−x=0⇔x2−x−1=0\sqrt{x^3-x^2-x+1}=1\Leftrightarrow x^3-x^2-x=0\Leftrightarrow x^2-x-1=0The solutions are x1.2=1±52x_{1.2}=\frac{1\pm\sqrt{5}}{2}But x∈[1,∞)x\in[1,\infty)so x=1+52x=\frac{1+\sqrt{5}}{2}
 
red12dog34 said:
x2−1x+x−1x=x⇔x2−1+x−1=xx⇔⇔x2+x−2+2x3−x2−x+1=x3⇔2x3−x2−x+1=(x3−x2−x+1)+1\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
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I actually had done exactly what you did but I failed to let t= x3−x2−x+1\displaystyle \sqrt{x^3-x^2-x+1}
In the corner again!
 
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