Question

find the zeros of the polynomial 9t2-6t+1

Answer 1

here is your answer!!

9t2-3t-3t+1
=3t(3t-1)-1(3t-1)
=(3t-1)(3t-1)
so the two zeroes are:
3t-1=0
t=1/3
the two zeroes are same.
i hope u get it.

Answer 2

HEYA MATE,

HERE IS UR ANSWER.

Thanks for asking this question.

9t² - 6t + 1 = 0
⇒ 9t² - 3t - 3t + 1 = 0
⇒ 3t(3t-1) -1(3t-1) = 0
⇒ (3t-1)(3t-1)=0
⇒ (3t-1)² = 0
So both roots are same
⇒ 3t = 1
⇒ t = 1/3

Verification of relationships:
Sum of roots = (1/3)+(1/3) = 2/3 = -(-6)/9 = -b/a
Product of roots = (1/3)*(1/3) = 1/9 = c/a

I HOPE IT HELPS U.