Question

If a and b are roots of equation x^2- x+1=0 then a^2009+b^2009 equalto

Answer 1

Answer is = - (a^2 + b^2) = 1.

x^2 - x + 1 = 0
Multiply with 1+x... So 1 + x^3 = 0.
So x = cube root of -1.
a^3 = -1. b^3 = -1.
Also from the quadratic equation we get
a+b = -1. ab = 1.

LHS = a^2009 + b^2009
= a^2 × (a^3)^669 + b^2 × (b^3)^669
= - a^2 - b^2
= - [ (a+b)^2 - 2ab ]
= - [ 1 - 2 ]
= 1

Answer 2

★ COMPLEX RESOLUTION ★

PROPERTIES OF CUBE ROOTS OF UNITY ARE BEING USED HERE , LITERALLY , ONLY FRAGMENTING ONE , REST IS ALGEBRAIC COMPUTATION

GIVEN EQUATION : x² - x + 1 = 0

And , x³ + 1 = x² - x + 1( x + 1) = 0

Given that a and b are the roots of the given equation , then , a and b will subsequently satisfy the original equation , aslike ,

here , product equivalent is also zero , hence , we may write the equation as

ƒ(x) = x + 1 ( x² - x + 1 ) = 0

a³ + 1 = a + 1 ( a² - a + 1 ) = 0

(a + 1) ( 0 ) = 0

a = -1

∴ a³ = -1
Similarly , b³ = -1

By General format of standard quadratic equation -

x²- x + 1 = ( x - a ) ( x - b ) = 0

x² - ( a + b ) x + ab = 0

a + b = 1 and ab = 1

a² + b² = -1

a²⁰⁰⁹ + b²⁰⁰⁹

a³ ⁽ ⁶⁶⁹ ⁾ ⁺ ² + b³ ⁽ ⁶⁶⁹ ⁾ ⁺ ²

a³⁽⁶⁶⁹⁾ a² + b³⁽⁶⁶⁹⁾ b²

(-1)⁶⁶⁹ a²+ (-1)⁶⁶⁹ b²

(-1)⁶⁶⁹ ( a² + b² )

( -1 ) ( -1 ) = 1

HENCE THE ANSWER IS 1

★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★