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 \( k = xy \). Then \( c = 16k + 49k = 65k \).

Rearranging the equation \( c = 65k \) into a quadratic form in \( k \) is not possible as it is a linear equation. The prompt seems to introduce a quadratic equation \( 16k^2 - ck + 49 = 0 \) for \( k \), which implies a misunderstanding or an unrelated problem statement in the original input. Assuming the intent is to analyze the condition for \( k \) to be real based on this quadratic equation:

Condition for Real Roots

For \( k \) to be real, the quadratic equation \( 16k^2 - ck + 49 = 0 \) must have real solutions. This requires the discriminant \( D \) to be non-negative: \[ D = b^2 - 4ac = c^2 - 4 \cdot 16 \cdot 49 \] \[ D = c^2 - 3136 \]

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

Conclusion

The discriminant must be non-negative for \( k \) to be real. Therefore, the condition is \( |c| \geq 56 \). Any value of \( c \) where \( |c|<56 \) will result in non-real values for \( k \).

Consequently, \( c = -50 \) is an invalid value because: \[ |-50| = 50<56 \] This violates the derived condition.

✅ Final Statement:
The value \( c = -50 \) is not permissible as it fails to meet the requirement \( |c| \geq 56 \).