Question

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

• A. -70
• B. 60
• C. -50
• D. -60

Answer 1

Answer: (B)

60

Answer 2

The given expression is $c = \frac{16x}{y} + \frac{49y}{x}$, where $x$ and $y$ are non-zero real numbers. We need to determine which of the given values $-70$, $60$, $-50$, and $-60$ cannot be taken by $c$. First, let's simplify the expression: $c = \frac{16x^2 + 49y^2}{xy}$ Now, we can rewrite the expression as: $c = \frac{16x^2}{xy} + \frac{49y^2}{xy} = 16\left(\frac{x}{y}\right) + 49\left(\frac{y}{x}\right) = 16k + \frac{49}{k}$ Where $k = \frac{x}{y}$. Since $x$ and $y$ are real and non-zero, $k$ is also real. We can now create a quadratic equation in terms of $k$: $16k^2 - ck + 49 = 0$ For $k$ to be real, the discriminant of this quadratic equation must be greater than or equal to $0$: $c^2 - 4 \cdot 16 \cdot 49 \geq 0$ $c^2 - 3136 \geq 0$ Solving for $c$, we get: $|c| \geq 56$ This means that the absolute value of $c$ must be greater than or equal to $56$. Therefore, $c$ cannot take the value $-50$, as $|-50| = 50 < 56$. So, out of the given options, $c$ cannot take the value $-50$.