2015 · Q23
For which value of k does the pair of equations x²-y²=0 and (x-k)²+y²=1 yield a unique positive solution of x?
Elementary MathematicsAlgebra
- A2
- B0
- C{2}
- D-{2}
Explanation
x² − y² = 0 gives y = ±x; substituting in (x−k)² + y² = 1 gives 2x² − 2kx + (k²−1) = 0. A unique (double) root needs discriminant 4k² − 8(k²−1) = 0, i.e. k² = 2. For k = √2 the root is x = √2/2 > 0 (unique and positive); k = √2 gives a negative root and k = 2 gives no real root. Hence k = √2 (option C). Stored key corrected from A to C.