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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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Answer :

TPS
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

kvnmurty
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.

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