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How can we show that $x^{8}+5x^{2}=1$ has exactly $2$ real roots?

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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.

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No problem - sorry if I sounded curt; I was in a rush – cats Jan 27 '13 at 23:19
No worries. Thanks for pointing out my mistake. – JavaMan Jan 28 '13 at 0:17
@lyj Thanks! Can you please explain what is "LHS", and also how I can show that there are exactly 2 roots, not more ? – Tina Jan 28 '13 at 6:55
LHS means "left-hand side"of the equation. – arsmath Jan 28 '13 at 21:11
@Tina, the fact that $f$ is strictly increasing on $[0,\infty)$ immediately gives that there are only two roots – cats Jan 29 '13 at 2:21

Hint: Descartes' rule of signs

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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).

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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.

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@Mike sure. I edited my post. – CBenni Jan 27 '13 at 19:43
Oops. My mistake, not yours. Probably would have noticed it had I done it on paper. – Mike Jan 27 '13 at 19:46

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 $f(x) = g(x)$.

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