Imaginary Square Roots

The trick here is to find out what is the range of values that $x$ can take.

For $y$ to be a real number, the expressions within the square roots must be non-negative, that is:

$\dfrac{6x+9}{7-0.69x} \geq 0 \quad$ and $\quad \dfrac{6x+9}{0.69x-7} \geq 0$

Now, let's solve it one by one.

For the left square root, let:

$6x+9 \geq 0 \implies x \geq -\dfrac{3}{2}\\ 7-0.69x > 0 \implies x < \dfrac{700}{69}$

Draw a number line or a table, and you will get that:

$\begin{cases} x < -\dfrac{3}{2} & \quad \dfrac{6x+9}{7-0.69x} < 0\\ -\dfrac{3}{2} \leq x < \dfrac{700}{69} & \quad \dfrac{6x+9}{7-0.69x} \geq 0\\ x > \dfrac{700}{69} & \quad \dfrac{6x+9}{7-0.69x} < 0\\ \end{cases}$

Therefore, the range of values of $x$ for the left square root is $-\dfrac{3}{2} \leq x < \dfrac{700}{69}$ or $\left[-\dfrac{3}{2},\dfrac{700}{69}\right)$ , whichever notation you prefer.

Repeat the same steps for the right square root (try it yourself!), and you should get $x \leq -\dfrac{3}{2},\; x > \dfrac{700}{69}$ or $\left(-\infty,-\dfrac{3}{2}\right] \cup \left(\dfrac{700}{69}, \infty\right)$ .

Now, what we need is values of $x$ that satisfies both ranges above. From these two ranges, we know that there is only one single exact value of $x$ that satisfies both ranges: $x=-\dfrac{3}{2}$

Therefore,

$y=\sqrt{\dfrac{6\left(-\frac{3}{2}\right)+9}{7-0.69\left(-\frac{3}{2}\right)}} + \sqrt{\dfrac{6\left(-\frac{3}{2}\right)+9}{0.69\left(-\frac{3}{2}\right)-7}} + 69 = \sqrt{0} + \sqrt{0} + 69 = \boxed{69}$

Since both $\frac{6x+9}{7-0.69x}$ and $\frac{6x+9}{0.69x-7}$ are nested under square roots, both are $\geq 0$ . WLOG suppose $6x+9\geq 0$ . For $\frac{6x+9}{7-0.69x}\geq 0$ , $7-0.69x\geq 0$ . But this implies $0.69x-7\leq 0\implies\frac{6x+9}{0.69x-7}\leq 0$ . But $\frac{6x+9}{0.69x-7}\geq 0$ since it is nested under a square root. Hence, both $\frac{6x+9}{7-0.69x}$ and $\frac{6x+9}{0.69x-7}$ are equal to $0$ , and our desired answer is $0+0+69=\boxed{69}$ .