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Find the zeroes of the quadratic polynomial 9t^2 - 6t + 1 and verify the relationship between the zeroes and coefficient.

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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

⇒ 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

Sum of roots = (1/3)+(1/3) = 2/3 = -(-6)/9 = -b/a

Product of roots = (1/3)*(1/3) = 1/9 = c/a

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

9 t² - 6 t + 1 = 0

t² - 2/3 t + 1/9 = 0

(t - 1/3)² = 0 t = 1/3 is a double root of the polynomial.

Verification of the relationship:

roots of ax² + b x + c = 0 are x = (-b+- √(b²-4ac) ) / 2 a

b = -6 a = 9 c = 1

zeros = roots = (6 + - √(6²-4*9) )/18 = (6 +- 0) / 18 = 1/3 so verified.

t² - 2/3 t + 1/9 = 0

(t - 1/3)² = 0 t = 1/3 is a double root of the polynomial.

Verification of the relationship:

roots of ax² + b x + c = 0 are x = (-b+- √(b²-4ac) ) / 2 a

b = -6 a = 9 c = 1

zeros = roots = (6 + - √(6²-4*9) )/18 = (6 +- 0) / 18 = 1/3 so verified.