Radicals and Quartic Equations

Squaring both sides of the equation, (keeping in mind that this method may yield extraneous solutions), we have

$x^{4} - 4x - 16 = 4 + x^{2} - 4x \Longrightarrow x^{4} - x^{2} - 20 = 0 \Longrightarrow (x^{2} - 5)(x^{2} + 4) = 0$ .

So either $x^{2} - 5 = 0 \Longrightarrow x = \pm \sqrt{5}$ or $x^{2} + 4 = 0 x = \pm 2i$ . As we are looking for real solutions we are left with $x = \sqrt{5}$ and $x = -\sqrt{5}$ as potential solutions. Now as $2 - \sqrt{5} \lt 0$ we can rule out $\sqrt{5}$ as a solution since the LHS of the equation is a square root, and hence (if real) must be non-negative. (Note that for $x = \sqrt{5}$ we have $\sqrt{x^{4} - 4x - 16} = \sqrt{9 - 4\sqrt{5}} = \sqrt{5} - 2 = -(2 - \sqrt{5})$ , explaining how this extraneous solution came about.)

Checking $x = -\sqrt{5}$ , we have

$\sqrt{x^{4} - 4x - 16} = \sqrt{9 + 4\sqrt{5}} = \sqrt{(2 + \sqrt{5})^{2}} = 2 + \sqrt{5} = 2 - x$ ,

Thus confirming that $x = -\sqrt{5}$ is the unique real solution. This gives us $a + b = -1 + 5 = \boxed{4}$ .

Solved by squaring both side. We will get 2 real and 2 imaginary solution. We will get sum of a + b either 4 or 6 , but if we think there is one more case. LHS of question is in root so it must be greater than zero. So by eliminating √5 (6) we can say answer is -√5 (4)