Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
Answer: x2 + y2 + z2 = (x + y + z)2 − 2(xy + yz + zx) = 0 − 0 = 0. A sum of squares of rational numbers is 0 only if each square is 0, so x = y = z = 0.
Step-by-step solution
Idea: Use the identity (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) to turn the two conditions into a statement about squares. Squares are never negative.
- (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx).1 mark
- Substitute the given values: 02 = x2 + y2 + z2 + 2 × 0, so x2 + y2 + z2 = 0.1 mark
- Each of x2, y2, z2 is ≥ 0. Three numbers that are each ≥ 0 can add to 0 only if each is 0. So x2 = y2 = z2 = 0, giving x = y = z = 0.1 mark
x² + y² + z² = (x + y + z)² − 2(xy + yz + zx) = 0, and since squares are non-negative, x = y = z = 0.
Check: If, say, x = 1, y = −1, z = 0, then x + y + z = 0 but xy + yz + zx = −1 ≠ 0, as the result predicts ✓.
Answer to write in the exam
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
0 = x2 + y2 + z2 + 0 ⇒ x2 + y2 + z2 = 0
x2, y2, z2 ≥ 0 ⇒ x2 = y2 = z2 = 0
∴ x = y = z = 0
Common mistakes that cost marks
- Trying a few numbers and concluding; examples cannot show that no non-zero solution exists.
- Writing (x + y + z)2 = x2 + y2 + z2 (forgetting the cross terms).
- Stopping at x2 + y2 + z2 = 0 without explaining why each must be 0.
How this can come in the exam
MCQ (1 mark)
If a + b + c = 6 and ab + bc + ca = 11, then a2 + b2 + c2 =
- 14
- 25
- 36
- 47
Show answer
(A) 14
36 − 2 × 11 = 14.
Short answer (2 marks)
If p + q = 0 and pq = 0 for rational p, q, show that p = q = 0.
Show answer
p2 + q2 = (p + q)2 − 2pq = 0 (1 mark). Squares are ≥ 0, so p2 = q2 = 0 and p = q = 0 (1 mark).Try one yourself
If x + y + z = 3 and xy + yz + zx = 3, find x2 + y2 + z2.
Show answer
9 − 6 = 3 (e.g. x = y = z = 1).
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