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

(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

(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

(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

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