Question

Show that the equation $x^4-12x^2+12x-3=0$ has a root between $-3$ and $-4$ and another between $2$ and $3$.

Answer 1

Given equation, $x^4-12x^2+12x-3=0$

Consider $f\left(x\right)=x^4-12x^2+12x-3$.

Then $f(-3)=-66$ and $f(-4)=13$

Since the signs of $f(-3)$ and $f(-4)$ are opposite, $f\left(x\right)$ must cross x-axis atleast once in the interval $(-4,-3)$.

$\therefore f\left(x\right)=0$ must have one root between $-3$ and $-4$.

Similarly, $f\left(2\right)=-11$ and $f\left(3\right)=6$.

Since the signs of $f\left(2\right)$ and $f\left(3\right)$ are different, $f\left(x\right)=0$ must have one root between $2$ and $3$.