This page Samacheer Kalvi Math Solution for Exercise 3.3 part 3 is going to provide you solution for every problems that you find in the exercise no 3.3
(vii) 2 x² - 2 √2x + 1
Solution:
Let p(x) = 2 x² - 2 √2x + 1 = (√2x - 1) (√2x - 1)
p(x) = 0 => (√2x - 1) (√2x - 1) = 0
√2x - 1 = 0
√2x = 1
x = 1/√2
x = 1/√2 x = 1/√2
p(1/√2) = 2 (1/√2)² - 2 √2(1/√2) + 1
= 1 - 2 + 1
= 2 - 2
= 0
sum of zeroes = (1/√2) + (1/√2) = 2/√2
product of zeroes = (1/√2) (1/√2) = 1/2
2 x² - 2 √2x + 1
ax² + bx + c
a = 2, b = -2 √2 and c = 1
Sum of roots α + β = -b/a
α + β = -(-2 √2)/2
α + β = 1
Product of roots α β = c/a
α β = 1/2
Thus the relationship verified
(viii) x² + 2 x - 143
Solution:
Let p(x) = x² + 2 x - 143 = (x - 11) (x + 13)
p(x) = 0 => (x - 11) (x + 13) = 0
x - 11 = 0 x + 13 = 0
x = 11 x = -13
x = 11 x = -13
p(11) = (11)² + 2 (11) - 143
= 121 + 22 - 143
= 143 - 143
= 0
p(-13) = (-13)² + 2 (-13) - 143
= 169 - 26 - 143
= 143 - 143
= 0
sum of zeroes = 11 + (-13) = -2
product of zeroes = 11(-13) = -143
x² + 2 x - 143
ax² + bx + c
a = 1, b = 2 and c = -143
Sum of roots α + β = -b/a
α + β = 2/1
α + β = 2
Product of roots α β = c/a
α β = -143/1
= -143
Thus the relationship verified
In the page samacheer kalvi math solution for exercise 3.3 part 3 we are going to see the solution of next problem
(2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 3 , 1
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 3 β = 1
x² - (3 + 1) x + 3 (1) = 0
x² - 4 x + 3 = 0
(ii) 2 , 4
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 2 β = 4
x² - (2 + 4) x + 2 (4) = 0
x² - 6 x + 8 = 0
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