Question

If \( x^2 + y^2 + z^2 = 3 \), which of the following cannot be the value of \( xy + yz + zx \)?

• A. -2
• B. -1
• C. 0
• D. 1
• E. 2

Hint:

In problems with quadratic expressions, expanding squares and using inequalities can help determine possible values for the unknowns.

Answer 1

The Correct Option is A

Step 1: Use the given equation.
We are given that \( x^2 + y^2 + z^2 = 3 \). We need to determine which of the following cannot be the value of \( xy + yz + zx \). Step 2: Analyze the possible values.
The expression \( (x + y + z)^2 \) can be expanded as: \[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx). \] Substitute \( x^2 + y^2 + z^2 = 3 \) into the equation: \[ (x + y + z)^2 = 3 + 2(xy + yz + zx). \] Let \( s = xy + yz + zx \), so the equation becomes: \[ (x + y + z)^2 = 3 + 2s. \] Step 3: Check the possible values for \( s \).
We know that \( (x + y + z)^2 \geq 0 \), because it is a square. Hence, \[ 3 + 2s \geq 0. \] This simplifies to: \[ 2s \geq -3, \] \[ s \geq -\frac{3}{2}. \] Step 4: Conclusion.
The value \( s = -2 \) is not possible, as it does not satisfy the inequality \( s \geq -\frac{3}{2} \). Therefore, the correct answer is \( -2 \).