SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 3.3 PART 1

This page Samacheer Kalvi Math Solution for Exercise 3.3 part 1 is going to provide you solution for every problems that you find in the exercise no 3.3

Samacheer Kalvi Math Solution for Exercise 3.3 part 1

(1) Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients

(i) x² - 2 x - 8 

Solution:

Let p(x) = x² - 2 x - 8 = (x - 4) (x + 2)

p(x) = 0 => (x - 4) (x + 2) = 0

          x = 4    x = - 2

p(4) = 4² - 2 (4) - 8

        = 16 - 8 - 8 

        = 16 - 16

        = 0

p(-2) = (-2)² - 2 (-2) - 8

         = 4 + 4 - 8

         = 8 - 8

         = 0

sum of zeroes = 4 + (-2) = 2

product of zeroes = 4 (-2) = -8

x² - 2 x - 8

ax² + bx + c

a = 1, b = -2 and c = -8

Sum of roots α + β = -b/a

                   α + β = -(-2)/1

                   α + β = 2

Product of roots α β = c/a

                      α β = -8/1

                      α β = -8

Thus the relationship verified

(ii) 4 x² - 4 x + 1

Solution:

Let p(x) = 4 x² - 4 x + 1 = (2x - 1) (2 x - 1)

p(x) = 0 => (2x - 1) (2 x - 1) = 0

      2 x - 1 = 0

       2 x = 1

          x = 1/2   x = 1/2

p(1/2) = 4 (1/2)² - 4 (1/2) + 1

        = 4/4 - 2 + 1

        = 2 - 2

        = 0

sum of zeroes = (1/2) + (1/2) = 1

product of zeroes = (1/2)(1/2) = 1/4

4 x² - 4 x + 1

ax² + bx + c

a = 4, b = -4 and c = 1

Sum of roots α + β = -b/a

                   α + β = -(-4)/4

                   α + β = 1

Product of roots α β = c/a

                      α β = 1/4

Thus the relationship verified

In the page samacheer kalvi math solution for exercise 3.3 part 1 we are going to see the solution of next problem

(iii) 6 x² - 3 - 7x

Solution:

Let p(x) = 6 x² - 7x - 3 = (2x - 3) (3 x + 1)

p(x) = 0 => (2x - 3) (3 x + 1) = 0

      2 x - 3 = 0

       2 x = 3

          x = 3/2

     3 x + 1 = 0

       3 x = -1

          x = -1/3

p (3/2) = 6 (3/2)² - 7(3/2) - 3

        = 6(9/4) - (21/2) - 3

        = (27/2) - (21/2) - 3

        = (27 - 21 - 6)/2

        = (27- 27 )/2

        = 0/2

        = 0

p (-1/3) = 6 (-1/3)² - 7(-1/3) - 3

        = 6(1/9) + (7/3) - 3

        = (2/3) + (7/3) - 3

        = (2 + 7 - 9)/2

        = (9 - 9 )/2

        = 0/2

        = 0

sum of zeroes = (3/2) + (-1/3) = (9 - 2)/3 = 7/3

product of zeroes = (3/2)(-1/3) = -1/2

6 x² - 7x - 3

ax² + bx + c

a = 6, b = -7 and c = -3

Sum of roots α + β = -b/a

                   α + β = -(-7)/6

                   α + β = 7/6

Product of roots α β = c/a

                      α β = -3/6

                           = -1/2

Thus the relationship verified