The numbers of pairs (a, b) of real numbers, such that whenever α is a root of the equation x² + ax + b = 0, α² - 2 is also a root of this equation, is :

Solution

Maths sir · 4 steps

The map x → x² - 2 must send each root back onto a root, so list every way that can happen.

1. Set up the map. Let the roots be α and β. The condition says α² - 2 and β² - 2 must each equal α or β. A root with r² - 2 = r maps to itself, and r² - r - 2 = 0 gives r = 2 or r = -1.
2. Swap case. If α² - 2 = β and β² - 2 = α with α ≠ β, subtracting gives α + β = -1. Adding gives α² + β² = 3, so 1 - 2αβ = 3 and αβ = -1.

   Pattern	Roots	Equation	(a, b)
   Equal roots, fixed	2, 2	x² - 4x + 4 = 0	(-4, 4)
   Equal roots, fixed	-1, -1	x² + 2x + 1 = 0	(2, 1)
   Both roots fixed	2, -1	x² - x - 2 = 0	(-1, -2)
   Roots swap	sum -1, product -1	x² + x - 1 = 0	(1, -1)
   β → 2, 2 → 2	2, -2	x² - 4 = 0	(0, -4)
   β → -1, -1 → -1	-1, 1	x² - 1 = 0	(0, -1)

3. Last pattern. In the last two rows both roots land on the same fixed root α. For α = 2, β² = 4 gives β = -2. For α = -1, β² = 1 gives β = 1.
4. So. Every pattern is covered and the six pairs are all different, so there are 6 pairs, option (a).

The idea to keep. When a map must send roots to roots, list the cases: repeated root, both roots fixed, roots swapped, and one root sent onto a fixed root. Find the fixed points first.

Answer from the NTA final answer key. Solution by PrepWiser.