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: Initial Setup.
We are given \( x^2 + y^2 + z^2 = 3 \) and must find which value *cannot* be \( xy + yz + zx \). Step 2: Equation Manipulation.
Expand \( (x + y + z)^2 \): \[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx). \] Substitute \( x^2 + y^2 + z^2 = 3 \): \[ (x + y + z)^2 = 3 + 2(xy + yz + zx). \] Let \( s = xy + yz + zx \): \[ (x + y + z)^2 = 3 + 2s. \] Step 3: Value Constraints.
Since \( (x + y + z)^2 \geq 0 \): \[ 3 + 2s \geq 0. \] Simplify: \[ 2s \geq -3, \] \[ s \geq -\frac{3}{2}. \] Step 4: Answer.
The value \( s = -2 \) is impossible because it violates \( s \geq -\frac{3}{2} \). The answer is \( -2 \).