Let α₁,α₂,....,α₇ be the roots of the equation x⁷ + 3x⁵ - 13x³ - 15x = 0 and |α ₁| ≥ |α ₂| ≥ ... ≥ |α ₇|. Then α₁α₂-α₃α₄+α₅α₆ is equal to _________.

Solution

Maths sir · 4 steps

Take out x, put t = x², and the degree-7 equation becomes a cubic you can factor at a glance.

1. Reduce to a cubic. Factor out x to get x(x⁶ + 3x⁴ - 13x² - 15) = 0. So one root is 0. Every other power is even, so put t = x² to get t³ + 3t² - 13t - 15 = 0.

   t³ + 3t² - 13t - 15 = (t+1)(t+5)(t-3)

2. Factor the cubic. t = -1 works, since -1 + 3 + 13 - 15 = 0. Dividing by (t+1) leaves t² + 2t - 15 = (t+5)(t-3). So x² = 3, -1 or -5.

   x²	Roots	Modulus	Product of the pair
   -5	± i√5 (α₁, α₂)	√5	(i√5)(-i√5) = 5
   3	±√3 (α₃, α₄)	√3	(√3)(-√3) = -3
   -1	± i (α₅, α₆)	1	(i)(-i) = 1
   0	0 (α₇)	0	not needed

3. Order does not matter. Inside each pair the two roots have the same modulus, so you cannot tell which one comes first. That does not matter, because the product of a pair ± r is always -r², whichever comes first.
4. So. α₁α₂ - α₃α₄ + α₅α₆ = 5 - (-3) + 1 = 9.

The common slip: 3. Taking α₃α₄ as +3 instead of -3 gives 5 - 3 + 1 = 3. The pair ±√3 multiplies to -3, and the minus sign in front then makes it +3.

The idea to keep. If a polynomial has only odd powers or only even powers of x, substitute t = x². Roots then come in pairs ±r, and each pair multiplies to −r², whatever the order.

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