An algebra problem by Mansi Bahuguna

$x+\frac{1}{x}=2$

or, $\frac{x^2+1}{x}=2$

or, $x^2-2x+1=0$

or, $(x-1)^2=0$

or, $x=1$

$\begin{aligned} x + \dfrac{1}{x} & = 2\\ \dfrac{x^2 + 1}{x} & = 2\\ x^2 + 1 & = 2x\\ x^2 - 2x + 1 & = 0\\ {(x - 1)}^2 & = 0\\ x - 1 & = 0\\ x & = \color{#69047E}{\boxed{1}} \end{aligned}$

We multiply both sides by x and we get x^2+x-2=0.So we use the quadratic formula to get x=(1+/-sqrt 1-(-8))/2.We evaluate....sqrt 1--8=3.(1+3)/2=2(answer).We can ignore the other solution which is extraneous and conclude x=2.

Look. X + 1 / X = 2, right? So, multiply both sides by x, we get that x+1 = 2x. So, subtracting 1 from both sides, we get that x = 1. And, we're done!

(x-1)^2=0 therefore x=1

$x+\frac{1}{x}=2$ Multiplying by $x$ on both sides,we get: $x^2+1=2x\rightarrow x^2-2x+1=0\rightarrow(x-1)^2=0\rightarrow x=1$

$x+1/x = 2$

$x^2+1=2x$

$x^2-2x+1=0$

$(x-1)^2=0$

$x=\boxed{1}$

(x+1/x=2)x 2x+1=2x

1

We have, x+1/x=2 or, x+1=2x or, x-2x=-1 or, -x=-1 so,x=1

We have, x + (1/x) = 2;
i.e x^2 + 1 = 2x;
i.e x^2 - 2x + 1 = 0;
i.e (x-1)^2 = 0;
i.e (x-1) = 0;
i.e x=1