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
(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
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