Question

find the zeros of the quadratic polynomial 9t²-6t+1 and verify the relationship between the zeros and the coefficient

Answer 1

9t²-6t+1
=9t²-3t-3t+1
=3t(3t-1)-1(3t-1)
=(3t-1)(3t-1)
t=1/3,1/3
product of zeroes = 1/3*1*3 = 1/9
sum of zeroes = 1/3+1/3 = 2/3
Also, product of zeroes = c/a = 1/9
sum of zeroes = -b/a = -(-6)/9 = 6/9 = 2/3
Hence verified.

Answer 2

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