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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