An algebra problem by Cj Lagmay

First note that $(\sqrt{x} + \sqrt{x + 7})^{2} = x + 2\sqrt{x^{2} + 7x} + x + 7 = 2x + 7 + 2\sqrt{x^{2} + 7x}$ .

So $2\sqrt{x^{2} + 7x} = (\sqrt{x} + \sqrt{x + 7})^{2} - (2x + 7)$ .

Substituting this into the given equation gives us that

$\sqrt{x} + ((\sqrt{x} + \sqrt{x + 7})^{2} - (2x + 7)) + \sqrt{x + 7} = 35 - 2x \Longrightarrow$

$(\sqrt{x} + \sqrt{x + 7}) + (\sqrt{x} + \sqrt{x + 7})^{2} = 35 - 2x + (2x + 7) = 42$ .

Now let $y = \sqrt{x} + \sqrt{x + 7}$ . This last equation can then be written as

$y + y^{2} = 42 \Longrightarrow y^{2} + y - 42 = 0 \Longrightarrow (y + 7)(y - 6) = 0$ .

Now as $y$ involves the sum of two square roots we know that $y \ge 0$ , so we choose the positive root $y = 6$ , and so $\sqrt{x} + \sqrt{x + 7} = 6$ . Squaring both sides yields

$x + 2\sqrt{x^{2} + 7x} + x + 7 = 36 \Longrightarrow 2\sqrt{x^{2} + 7x} = 29 - 2x$ .

Squaring both sides yet again gives us that

$4(x^{2} + 7x) = 29^{2} - 116x + 4x^{2} \Longrightarrow 28x + 116x = 841 \Longrightarrow x = \dfrac{841}{144} = \left(\dfrac{29}{12}\right)^{2}$ .

Thus $a + b = 841 + 144 = \boxed{985}$ .

It can be checked that this value of $x$ does indeed satisfy the original equation. Also, we know that it is the only real solution since the expression on the left of the original equation is a continuously increasing function starting at $(0, \sqrt{7})$ , and the expression on the right is a continuously decreasing function intersecting the y-axis at $(0,35)$ and the x-axis at $\left(\dfrac{35}{2}, 0\right)$ .