Show that the equation $2x^4-9x^2+4 = 0$ has at least one solution in $(0,1)$ Show that the equation $2x^4-9x^2+4 = 0$ has at least one solution in $(0,1)$.
It is not possible to show it by Bolzano's theorem because neither 0 nor 1 are in the given interval, is it? Is there any way to do it other than solving it agebraically or analysing its graph? By the way, the root in $(0, 1)$ is $x = \frac{1}{\sqrt2}$.
Thanks in advance.
 A: What about the fact that $f(0) > 0$, and $f(1) < 0$, so because the function is continuous, the function must take on all y-values in between $f(0)$ and $f(1)$. This includes $0$.
A: You can explicitly solve it to get
$\pm\sqrt{(9\pm 7)/4}
=\pm\sqrt{4, 1/2}
=\pm 2,\pm 1/\sqrt{2}$
and the smaller positive root
is between 0 and 1.
A: You could still use Bolzano's theorem. Although the endpoints are not within the interval, due to the continuity of the function if the endpoints have opposite signs the theorem will still hold.
The endpoints are $(0,4)$ and $(1,-3)$. These can be easily calculated. Since, as RayDansh mentioned, the function must take on all $y$-values from -3 to 4, the function must have a zero somewhere within the given interval by the intermediate value theorem. It does not matter that the endpoints are not within the interval as the function is continuous.
A: For a general equation, you have to apply the intermediate value theorem. But, luckly, for your concrete equation, you may directly solve it.
Since $$2x^4-9x^2+4 =(2x^2-1)(x^2-4)=(\sqrt{2}x+1)(\sqrt{2}x-1)(x+2)(x-2)= 0.$$
Thus, $$x_1=-\frac{\sqrt{2}}{2},~~~x_2=\frac{\sqrt{2}}{2},~~~x_3=-2,~~~x_4=2.$$
It's clear that $$x_2 \in (0,1).$$
