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. -50
• C. 60
• D. -60

Answer 1

The Correct Option is B

Given: \[ c = 16xy + 49yx \]

Let: \[ k = xy \Rightarrow c = 16k + 49k = 65k \]

Now, rearranging to express as a quadratic in \( k \): \[ 16k^2 - ck + 49 = 0 \]

Condition for Real Roots

For \( k \) to be real, the quadratic must have a real solution. That is, the discriminant \( D \) must be non-negative: \[ D = b^2 - 4ac = c^2 - 4 \cdot 16 \cdot 49 \] \[ D = c^2 - 3136 \]

For \( D \geq 0 \), we need: \[ c^2 \geq 3136 \Rightarrow |c| \geq \sqrt{3136} = 56 \]

Conclusion

Since the discriminant must be non-negative, we conclude: \[ |c| \geq 56 \] Therefore, any value of \( c \) such that \( |c| < 56 \) would make \( k \) non-real.

Hence, \( c = -50 \) is not valid as: \[ |-50| = 50 < 56 \] So it violates the condition.

✅ Final Statement: 
The value \( c = -50 \) is not allowed because it does not satisfy the requirement \( |c| \geq 56 \).