Real Solutions,

**craigmain** · March 28th 2009, 06:04 AM

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.

**earboth** · March 28th 2009, 06:11 AM

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.

**HallsofIvy** · March 28th 2009, 06:18 AM

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.

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