# Real Solutions,

• Mar 28th 2009, 06:04 AM
craigmain
Real Solutions,
Hi,

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

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.

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

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.
• Mar 28th 2009, 06:11 AM
earboth
Quote:

Originally Posted by craigmain
Hi,

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

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.

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

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.

$\displaystyle \frac{2}{x^2}+x^4=5~\implies~x^6-5x^2+2=0$

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

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

You now can apply the Cardanic formula to solve this equation. Don't forget to re-substitute to calculate the value of x.
• Mar 28th 2009, 06:18 AM
HallsofIvy
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, $\displaystyle x^6- 5x^2= 2$ or $\displaystyle x^6- 5x^2- 2= 0$. If you let $\displaystyle y= x^2$ you can write that as $\displaystyle y^4- 5y- 2= 0$, a cubic equation. By the "rational root" theorem, the only possible rational roots are $\displaystyle \pm 1$ or $\displaystyle \pm 2$, 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.