Simply roots of equation

Since the roots are symmetric around $0$ , they must be $\pm a, \pm 3a$ for some $a$ . So we get $(x-3a)(x-a)(x+a)(x+3a) = (x^2-1)(x^2-4)-k$ Equating coefficients of $x^2$ on both sides gives $-10a^2 = -5$ , or $a^2 = 1/2$ . Equating constant terms gives $9a^4 = 4-k$ , so $k = 4-9a^4 = 4-9/4 = \fbox{1.75}$ .