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

The Correct Option is B

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 \(-50\).