a line and a polynomial

For $y = x^4-2x^2 - ax -b$ to have four distinct roots, it must have three turning points. Turning points occur when $\dfrac {dy}{dx} = 0$ . In this case, $\dfrac {dy}{dx} = 4x^3-4x - a = 0$ . The curve $f'(x)$ is "S" shape as shown in the figure. The red curve is with $a=0$ . We note that as $a$ increases, the curve shifts down and when $a > y_{\text{max}}$ the local maximum, $\dfrac {dy}{dx}$ has only one zero, and hence $y(x)$ has only one turning point and have two solutions. Therefore the $a_{\text{max}} = y_{\text{max}}$ . To find $y_{\text{max}}$ , find $x$ when $\dfrac {d^2y}{dx^2} = 0$ or $12x^2 - 4 = 0\implies x = - \dfrac 1{\sqrt 3}$ and $a_{\text{max}} = y_{\text{max}} = - \dfrac 4{3\sqrt 3} + \dfrac 4{\sqrt 3} = \dfrac 8{3\sqrt 3} \approx \boxed{1.540}$ .

The given polynomial will have four distinct real roots if all the roots of it's first derivative with respect to it's argument are real. That is, all the roots of the equation $4x^3-4x-a=0$ are real. This will be true when $-27\times 4^2\times (a) ^2-4\times 4\times (-4)^3\geq 0$ or $27a^2\leq 64$ or $a\leq \dfrac{8}{3\sqrt 3}$ . Hence, the maximum value of $a$ is $\dfrac{8}{3\sqrt 3}\approx \boxed {1.5396}$