If $C$ = $\frac{16x}{y} + \frac{49y}{x}$ for some non-zero real numbers $x$ and $y$, then $c$ cannot take the value

Given: $C = \frac{16x}{y} + \frac{49y}{x}$ where $x$ and $y$ are non-zero real numbers.

We'll use a substitution technique to convert this into a quadratic equation, then use the discriminant to find which values are impossible.

---

Let's set $s = \frac{x}{y}$. This substitution helps because:

Since $x$ and $y$ are non-zero real numbers, $s$ is also a non-zero real number

We can rewrite our expression in terms of $s$

Rewriting $C$:

$C = \frac{16x}{y} + \frac{49y}{x} = 16 \cdot \frac{x}{y} + 49 \cdot \frac{y}{x} = 16s + \frac{49}{s}$

---

To eliminate the fraction, we multiply both sides by $s$:

$Cs = 16s^2 + 49$

$16s^2 - Cs + 49 = 0$

Key Insight: Since $s = \frac{x}{y}$ is a real number (as $x$ and $y$ are real), this quadratic equation must have real solutions.

---

For a quadratic equation $as^2 + bs + c = 0$, the discriminant is $D = b^2 - 4ac$. The equation has real solutions only when $D \geq 0$.

For our equation $16s^2 - Cs + 49 = 0$:

$a = 16$

$b = -C$

$c = 49$

Since we need real solutions:

$D = (-C)^2 - 4(16)(49) \geq 0$

$C^2 - 4 \times 16 \times 49 \geq 0$

$C^2 - 3136 \geq 0$

---

$C^2 \geq 3136$

Taking square roots (remember both positive and negative roots):

$|C| \geq \sqrt{3136} = 56$

This means: $C \leq -56$ or $C \geq 56$

---

The values C cannot take: $-56 < C < 56$

We can see that -50 lies in this forbidden range since $-56 < -50 < 56$.

Therefore, C cannot take the value -50.

---

Alternative approach: For positive terms, we can use AM-GM inequality:

$\frac{16x}{y} + \frac{49y}{x} \geq 2\sqrt{\frac{16x}{y} \cdot \frac{49y}{x}} = 2\sqrt{16 \times 49} = 2 \times 56 = 112$

However, since $x$ and $y$ can have different signs, we need to consider negative cases too, leading to the same conclusion: $C \leq -56$ or $C \geq 56$.