1.

Find the zeroes of the quadratic polynomial 9x2 – 6x + 1 and verify the relationship between the zeroes and the coefficients​

Answer»

Step-by-step explanation:

GIVEN:-

The quadratic polynomial 9x^2 – 6x + 1

To FIND:-

Find the zeroes of the quadratic polynomial

9x^2 – 6x + 1 and VERIFY the relationship between the zeroes and the coefficients

Solution:-

Given quadratic polynomial is 9x^2-6x+1

=>P(x)= 9x^2 - 6x +1

=>P(x)= 9x^2 -3x -3x +1

=>P(x) = 3x ( 3x -1) -1( 3x-1)

=>P(x) = (3x - 1) (3x-1)

To get the zeores we can write P(x) = 0

=> P(x) = (3x - 1) (3x-1) = 0

=> 3x -1 = 0

=> 3x = 1

=> x = 1/3

The zeores are 1/3 and 1/3

Let α = 1/3 and β = 1/3

P(x)=9x^2 - 6x +1

On Comparing this with the standard quadratic Polynomial ax^2+bx+c

a = 9

b=-6

c=1

Sum of the zeros =

α + β = (1/3)+(1/3)

=> (1+1)/3

=> 2/3

=-(-6/9)

=> -b/a

α + β = b/a

Product of the zeroes = α β

α + β = (1/3)(1/3)

=> 1/9

=> c/a

α β = c/a

Verified the given relations between the zeroes and the coefficients of the given Polynomial.

Used formulae:-

  • Sum of the zeros = -b/a

  • Product of the zeroes = c/a


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