SAMACHEER KALVI MATH SOLUTION FOR EXERCISE 3.3 PART 2

This page Samacheer Kalvi Math Solution for Exercise 3.3 part 2 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 2

(iv) 4 x² + 8 x

Solution:

Let p(x) = 4 x² + 8 x = 4 x (x + 2)

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

          4 x = 0    x  = - 2

            x = 0     x = -2

p(0) = 4 (0)² + 8 (0)

        = 0 + 0

        =  0

p(-2) = 4 (-2)² + 8 (-2)

         = 4(4) - 16

        = 16 - 16

        =  0

sum of zeroes = 0 + (-2) = -2

product of zeroes = 0 (-2) = 0

4 x² + 8 x

ax² + bx + c

a = 4, b = 8 and c = 0

Sum of roots α + β = -b/a

                   α + β = (-8)/4

                   α + β = -2

Product of roots α β = c/a

                      α β = 0/4

                      α β = 0

Thus the relationship verified

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(v)  x² - 15

Solution:

Let p(x) = x² - 15 = x² - (√15)²

                          = (x + √15) (x - √15)

p(x) = 0 => (x + √15) (x - √15) = 0

      (x + √15) = 0          (x - √15) = 0

        x = - √15                x = √15

p(√15) = (√15)² - 15

        = 15 - 15

        = 0

p(-√15) = (-√15)² - 15

        = 15 - 15

        = 0

sum of zeroes = (-√15  ) + ( √15  ) = 0

product of zeroes = (- √15 )( √15  ) = -15

x² - 15

ax² + bx + c

a = 1, b = 0 and c = -15

Sum of roots α + β = -b/a

                   α + β = 0/1

                   α + β = 0

Product of roots α β = c/a

                      α β = -15/1

                            = -15  

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

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(vi)  3 x² - 5 x + 2

Solution:

Let p(x) =  3 x² - 5 x + 2 = (3x - 2) (x - 1)

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

      3 x - 2 = 0

       3 x = 2

          x = 2/3

     x - 1 = 0

       x = 1

p (2/3) = 3 (2/3)² - 5 (2/3) + 2

        = 3(4/9) - (10/3) + 2

        = (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)/6 = 7/6

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