Basic Math Problem of the Week 11/24/2017

[PF PotW Robot]

Here is this week's basic math problem. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Using spoiler tags is optional. Occasionally there will be prizes for extraordinary or clever methods.

Solve $\frac{1}{4}\left(\sin\left(\frac{\pi x}{2}\right)\right)^2+2x^4-5x^2+1=0$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[.Scott]

For those who don't want to think too much, we know that the $-5x^2$ term is the only one that can offset all the other positive terms. So x needs to be well over zero but not enough to make the $+4x^4$ too large. So setting this up in a spreadsheet with values of x from 0 to 1.5 quickly reveals the answer.

 

[fresh_42]

W.l.o.g.

So x needs to be well over zero

since the function is symmetric to $x=0$.

 

[Charles Link]

I got lucky and found one root immediately. $x=\pm (1/2)$. $\\$ Editing: Making a quadratic equation out of the fourth power equation with the value at $x=\pm (1/2)$ inserted for $\sin^2(\frac{\pi x}{2})=\frac{1}{2}$ shows that $x=\pm (1/2)$ is not a double root of the fourth power equation= quadratic equation in $x^2$. Instead, the factoring is $(2x^2-\frac{1}{2})(x^2-\frac{9}{4})=0$. Trying $x=\pm (3/2)$, (solution to $x^2-\frac{9}{4}=0$), gives the same value for $\sin^2(\frac{\pi x}{2})=\frac{1}{2}$ that $x=\pm (1/2)$ gave. Thereby, the other roots are in fact $x=\pm 3/2$.

 

[Charles Link]

Please see my edited addition to post 4.