Show that $x^{8}+5x^{2}=1$ has exactly $2$ real roots How can we show that $x^{8}+5x^{2}=1$ has exactly $2$ real roots?
 A: Hint: Descartes' rule of signs
A: Let $f(x)=x^8+5x^2-1$. 
If $f$ has at least $3$ real roots then $f''(x)=56x^6+10$ will have at least $1$ real root ↯.
Since $f(0)<0<f(1), \ f$ has exactly two real roots (polynomial of even degree). 
A: How about this: Differentiate. You see that $f'(x)$ has only one root $(x=0)$ and therefore, there is a maximum of 1 extremal point. As f is continuous, we calculate $f(-5)$, $f(0)$ and $f(5)$ and use the intermediate value theorem to prove there are exactly 2 real roots.
A: Note that the left-hand side is even and strictly increasing on $[0, \infty).$ If we let $f(x) = LHS,$ then $f(0) = 0$ and $f(1) = 6,$ so by the intermediate value theorem, there exists $x\in(0,1)$ such that $f(x) = 1.$ Then $f(-x) = 1$ as well, and those are your two roots.
A: Let $f(x)= x^8$ and $g(x)= 1- 5x^2$, let's analyze the solutions of $x^8=1-5x^2$, i.e . $f(x)=g(x)$.
See the graphs below:

$f(x)$ is always concave upwards , and $g(x)$ is always concave downwards.(Check the second derivatives of both functions). As $f(0)=0$ (global minimum of f(x)) , $g(0) = 1$ (global maximum of g(x)) and both functions are even and continuous there will be two solutions to $f(x) = g(x)$.
