# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.23.3" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Consider the two curves $y=2x^3+6x+1$ and $y=-3/x^2$ in the Cartesian plane. Find the number of distinct points at which these two curves intersect.

#### Singapore Math Olympiad 2006 (Senior Section - Problem 23)

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" custom_padding="|||22px||"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]Try to think how to find the intersection points using these two given equation $y=2x^3+6x+1$ and $y=-3/x^{2}$

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Try to compare the two values of y e.g.$2x^3+6x+1=-3/x^{2}$ Do we get two factors $2x^3+1=0$ and $x^2+3=0$

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]

At end as we will consider only $2x^3+1=0$ as $x^2+3>0$

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

Thus we will get the value of x and from there we can find the value of y and we will get the answer.  $(\frac{-1}{\sqrt [3]{2}}, \frac{-3}{\sqrt [3]{4}})$

So the number distinct point will be 1