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

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Accepted Answer

#### Step 1: Rewrite the Expression for $$ c $$:  
Let $$ k = \frac{x}{y} $$. Since $$ x $$ and $$ y $$ are real and non-zero, $$ k $$ is also real and non-zero.  
Substitute into the expression for $$ c $$:  
$$
c = 16k + \frac{49}{k}.
$$

Multiply through by $$ k $$ to eliminate the fraction:  
$$
ck = 16k^2 + 49.
$$

Rearrange into a quadratic equation:  
$$
16k^2 - ck + 49 = 0.
$$

#### Step 2: Conditions for Real Values of $$ k $$:  
For $$ k $$ to be real, the discriminant of the quadratic equation must be non-negative:  
$$
\Delta = (-c)^2 - 4(16)(49) \geq 0.
$$

Simplify the discriminant:  
$$
c^2 - 4(16)(49) \geq 0.
$$

Calculate the constant term:  
$$
c^2 - 3136 \geq 0.
$$

This implies:  
$$
|c| \geq \sqrt{3136} = 56.
$$

#### Step 3: Identify Values That $$ c $$ Cannot Take:  
From the condition $$ |c| \geq 56 $$, any value of $$ c $$ with absolute magnitude less than 56 is not possible. Specifically, $$ c = -50 $$ violates this condition because $$ |-50| < 56 $$.

#### Final Answer:  
The value that $$ c $$ cannot take is **-50**, and the correct answer is **C**.

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