1. ## Real Solutions,

Hi,

The graph $y=\frac{2}{x^2}+x^4$ has four real number solutions for values of $y > 3$.

I can draw the graph using my math tools, but this is quite a stinker to try and graph by hand.

I was wondering how you would find the four real solutions for a particular value of y, say 5. I have tried factoring it.

$\frac{2}{x^2}+x^4=5$
$x^2(x^4-5)=2$

There doesn't appear to be any way of factoring out the four real number solutions. What is the correct way to find values of the graph.

Thanks
Regards
Craig.

2. Originally Posted by craigmain
Hi,

The graph $y=\frac{2}{x^2}+x^4$ has four real number solutions for values of $y > 3$.

I can draw the graph using my math tools, but this is quite a stinker to try and graph by hand.

I was wondering how you would find the four real solutions for a particular value of y, say 5. I have tried factoring it.

$\frac{2}{x^2}+x^4=5$
$x^2(x^4-5)=2$

There doesn't appear to be any way of factoring out the four real number solutions. What is the correct way to find values of the graph.

Thanks
Regards
Craig.
$\frac{2}{x^2}+x^4=5~\implies~x^6-5x^2+2=0$

If you use the substitution $x^2 = y$ you'll get a reduced cubic equation:

$y^3-5y+2=0$

You now can apply the Cardanic formula to solve this equation. Don't forget to re-substitute to calculate the value of x.

3. The fact that an equation has 4 solutions does not mean that those solutions are rational numbers or that they can be found easily!

In this case, multiplying the left side, $x^6- 5x^2= 2$ or $x^6- 5x^2- 2= 0$. If you let $y= x^2$ you can write that as $y^4- 5y- 2= 0$, a cubic equation. By the "rational root" theorem, the only possible rational roots are $\pm 1$ or $\pm 2$, and it is easy to check that none of those satisfies the equation- thus the original equation has no rational roots.

You could use Cardan's cubic formula
Cubic function - Wikipedia, the free encyclopedia
to solve the cubic, then take the square roots of that.