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.
On
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 $$(x^2-x-1)^2=0$$
On
Hint:take $$a=x+\frac1x ,b=x-\frac1x$$ so $$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x\\\sqrt{a}+\sqrt{b}=\frac{a+b}{2}$$
On
$$\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2} = \frac{\left(x-\frac{1}{x}\right)-\left(1-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}}=\frac{x-1}{x}=1-\frac1x,$$ 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 $x>0$.
$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x\iff \left(x-\frac{1}{x}\right)^{1/2} = x-\left(1-\frac{1}{x}\right)^{1/2}.$$ 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.