Question

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

Answer 1

Hey!!!!

Good Evening

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We have

=> 9t² - 6t + 1 = 0

=> 9t² - 3t - 3t + 1 = 0

=> 3t(3t - 1) - (3t - 1) = 0

=> (3t - 1)² = 0

Thus t = 1/3 and t = 1/3 (equal roots)

$let \: \alpha = \beta = \frac{1}{3}$

Verification ⤵️

$1. \: \alpha + \beta = \frac{ - b}{a}$

Replacing value

$= > \frac{1}{3} + \frac{1}{3} = \frac{6}{9}$

$= > \frac{2}{3} = \frac{2}{3}$
Hence Verified

$2. \: \alpha \beta = \frac{c}{a}$

Replacing the values

$= > \frac{1}{3 } \times \frac{1}{3} = \frac{1}{9}$

$= > \frac{1}{9} = \frac{1}{9}$

Hence Verified

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Hope this helps ✌️

Answer 2

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Q. Find the zeroes of the quadratic polynomial  9 t² - 6 t + 1 and verify the relationship between zeroes and coefficient.


Solution : 9 t² - 6 t + 1

             = 9 t² - 3 t - 3 t + 1

            = 3 t ( 3 t - 1 ) - 1 ( 3 t - 1 )

            ( 3 t - 1 ) ( 3 t - 1 ) = 0

            ( 3 t - 1 ) = 0 / ( 3 t - 1 )  or ( 3 t - 1 ) = 0 / ( 3 t - 1 )

           ( 3 t - 1 ) = 0                   or ( 3 t - 1 ) = 0

              t = 1 / 3.                       or t = 1/ 3.

Sum of zeroes = - coefficient of x / coefficient of x²

1 / 3 + 1 / 3 = - ( - 6 ) / 9 

1 / 3 ( 1 + 1 ) = 6 / 9 

1 x 2 / 3 = 2 / 3.

2 / 3 = 2 /3 .Verified.


Product of zeroes = constant term / coefficient of x²

( 1 / 3 ) x ( 1 / 3 ) = 1 / 9

1 / 9 = 1 / 9.Verified.

The zeroes of ( 9 t²2 - 6 t + 1 ) are 1 / 3 and 1 / 3.



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