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How do I solve

$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x$$

My Try I tried to take the first term as $t$ but then I had to square both sides twice and that led to a complex bi quadratic. I'm not sure even that'll solve the problem.

guys please let me know what are different possible ways to tackle problems like these, and how would you go about on solving this particular one.

Arthur
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Tanuj
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4 Answers4

4

$$\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2} = \frac{\left(x-\frac{1}{x}\right)-\left(1-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}}=\frac{x-1}{x}=1-\frac1x,$$ adding both equations gives $$2\left(x-\frac{1}{x}\right)^{1/2}=x-\frac{1}{x}+1.$$ That means $$x-\frac{1}{x}=1,$$ so we have $x=\phi$ (the Golden Ratio). The other solution $x=-\phi^{-1}$ does not satisfy the original equation, we must have $x>0$.

1

$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x\iff \left(x-\frac{1}{x}\right)^{1/2} = x-\left(1-\frac{1}{x}\right)^{1/2}.$$ Squaring

$$x-\frac{1}{x} = x^2+1-\frac{1}{x}-2x\left(1-\frac{1}{x}\right)^{1/2}.$$ Simplifying

$$x = x^2+1-2x\left(1-\frac{1}{x}\right)^{1/2}.$$ That is

$$2x\left(1-\frac{1}{x}\right)^{1/2}=x^2-x+1.$$ Squaring again

$$4x^2\left(1-\frac{1}{x}\right)=x^4-2x^3+3x^2-2x+1.$$ Thus, we get

$$x^4-2x^3-x^2+2x+1=0.$$ That is

$$(x^2-x-1)^2=0.$$ This must be easy to solve.

mfl
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  • At last, how did you figure out the factors for the degree 4 expression? – Tanuj Aug 15 '17 at 10:03
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    $\left(x^2+ax+b\right)^2=x^4+2ax^3+a^2 x^2+2 b x^2+2 a b x+b^2\quad$ Now compare with the coefficient of $x^4-2x^3-x^2+2x+1\quad$ We have $a=-1$ and $x^4-2x^3+(2 b+1) x^2-2 b x+b^2$ and we have $2b+1=-1\to b=-1$ so the polynomial is $\left(x^2-x-1\right)^2\quad$ be careful because one of the solution is negative and is a fake solution – Raffaele Aug 15 '17 at 19:47
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    And, since we have squared, we might have introduced solutions, so don't forget to test the solutions found for the last equation in the original equation. – md2perpe Aug 17 '17 at 09:22
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HINT: square one times you will get $$x-\frac{1}{x}+1-\frac{1}{x}+2\left(x-\frac{1}{x}\right)^{1/2}\left(1-\frac{1}{x}\right)^{1/2}=x^2$$ rearranging $$2\left(x-\frac{1}{x}\right)^{1/2}\left(1-\frac{1}{x}\right)^{1/2}=x^2-x-\frac{2}{x}-1$$ an then you must square again after squaring and sorting the Terms we get $$(x^2-x-1)^2=0$$

Dr. Sonnhard Graubner
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Hint:take $$a=x+\frac1x ,b=x-\frac1x$$ so $$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x\\\sqrt{a}+\sqrt{b}=\frac{a+b}{2}$$

roshanaie
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