Question

guys class11 quadratic equation ml khanna question please help​

Answer 1

Answer:

(b)

Step-by-step explanation:

8x^2-10x+3=0

split the middle term

8x^2-6x-4x+3=0

we get, 1/2, 3/4 as roots

therefore,

alpha=1/2

beta=squareroot3/2

now use de-moivre's theorem

z^n=r^n(cosntheta) + isin(ntheta)

hence,

(1/2-iroot3/2)^100 and (1/2+iroot3/2)^100 are the two roots

now,

here, r= square root((1/2)^2+(root3/2)^2)

          =1

and, tantheta=(root3/2)/(1/2)

                     =root3

hence, theta= 60 degrees

therefore, cos60=1/2

  and        sin60=roo3/2

now, (1/2-iroot3/2)^100=((1)(cos60+isin60))^100=

1^100(cos 100*60 + isin 100*60)

=cos6000 + isin6000

=cos(180*33 + 60) + isin(180*33 + 60)

=-cos60-isin60

=-1/2-iroot3/2

similarly other is

-1/2+iroot3/2

these two roots satisfy only solution is x^2+x+1=0