Question

If alpha and beeta the roots of the Quadratic equation x²-2x+2=0 then the possible value of n for which (alpha/beeta) ^n=1​

Answer 1

Answer:

n= 4k where k is an integer

Step-by-step explanation:

now a,b are roots of the eqn

a+b = -(-2)/1 = 2

ab = 2

b= 2/a

a+2/a = 2

a2 -2a + 2 = 0

a = 2+_√4-8)/2

= 1+_i

b = 2-a

= 1_+i

thus solution to eqn are

If a= 1+i , b=1-i and if a= 1-i, b= 1+i

now

case 1 a= 1+I, b = 1-i

a/b = (1+i)/(1-i)

rationalising we get

a/b = (1+I)^2/(1-i^2)

= (1+i^2+2i)/(1-(-1))

= ((1-1+2i)/(1+1)

= 2i/2

= i

now any power of i in multiple of 4 is 1

hence one solution is n = 4k where k is an integer

case 2 a= 1-i,b=1+i

a/b = (1-i)/(1+I)

rationalising we get

a/b = (1-i)^2/(1-i^2)

= (1-1-2i)/(1+1)

a/b = (-i)

in this case also, solution will be n = 4k where k is an integer.