Question

What are the real or imaginary solutions of each polynomial equation?
X4 - 40x2 + 144 = 0

Answer 1

In order to solve equations like

$ax^4+bx^2+c=0$

we have to set up the auxiliary variable $t=x^2$

and the equation becomes a standard quadratic equation.

In your case, we have

$x^4-40x^2+144=0 \iff t^2-40t+144=0$

The solutions to this equation is given by

$t_{1,2}=\dfrac{40\pm\sqrt{1600-576}}{2}=\dfrac{40\pm\sqrt{1024}}{2}=\dfrac{40\pm 32}{2}=20\pm 16$

So, the two solutions are

$t_1=20+16=36,\quad t_2=20-16=4$

Now recall that $t=x^2$ to solve for $x$:

$t=36 \implies x^2=36 \iff x=\pm 6,\quad t=4 \implies x^2=4 \iff x=\pm 2$

So, the four solutions are

$x_1=-4,\quad x_2=-2,\quad x_3=2,\quad x_4=4$