Respuesta :

Try this:
1) If y=x² and y=(2x+1)/x, then 
[tex]x^2= \frac{2x+1}{x}; \ \ \textless \ =\ \textgreater \ \ x^3-2x-1=0; \ (x \neq 0)[/tex] <=>   [tex](x+1)(x^2-x-1)=0; \ (x \neq 0)\ \ \textless \ =\ \textgreater \ \left[\begin{array}{ccc}x=-1\\x= \frac{1+ \sqrt{5} }{2} \\x= \frac{1- \sqrt{5}}{2} \end{array}\right[/tex]
2) after 'x' are found it need to calculate 'y':
Answer: (-1;1), (-0.618;0.382), (1.618;2.618)

Answer:

[tex](-1,1),(-0.618,0.382), (1.618,2.618)[/tex]

Step-by-step explanation:

we have

[tex]y=x^{2}[/tex] ----> equation A

[tex]y=\frac{1}{x} +2[/tex] -----> equation B

we know that

The intersection points both graphs are the points where both equations have the same value

so

Using a graphing tool

The intersection points are [tex](-1,1),(-0.618,0.382), (1.618,2.618)[/tex]

see the attached figure


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