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)
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
