This page Samacheer Kalvi Math Solution for Exercise 3.3 part 4 is going to provide you solution for every problems that you find in the exercise no 3.3
(iii) 0,4
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 0 β = 4
x² - (0 + 4) x + 0 (4) = 0
x² - 4 x = 0
(iv) √2,1/5
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = √2 β = 1/5
x² - (√2 + (1/5)) x + √2 (1/5) = 0
x² - (5√2 + 1)/5 x + (√2/5) = 0
In the page samacheer kalvi math solution for exercise 3.3 part 4 we are going to see the solution of next problem
(v) 1/3,1
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 1/3 β = 1
x² - ((1/3) + 1) x + (1/3) (1) = 0
x² - [(1 + 3)/3] x + (1/3) = 0
x² - (4/3) x + (1/3) = 0
3x² - 4 x + 1 = 0
(vi) 1/2, -4
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 1/2 β = -4
x² - ((1/2) + (-4)) x + (1/2) (-4) = 0
x² - [(1 - 8)/2] x + (1/3) = 0
x² - (4/3) x + (1/3) = 0
3x² - 4 x + 1 = 0
(vii) 1/3,1/3
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = 1/3 β = 1/3
x² - ((1/3) + (1/3)) x + (1/3) (1/3) = 0
x² - [(1 + 1)/3] x + (1/9) = 0
x² - (2/3) x + (1/9) = 0
9x² - 6 x + 1 = 0
(viii) √3 , 2
Solution:
General form of quadratic equation with roots α and β is
x² - (α + β) x + αβ = 0
α = √3 β = 2
x² - (√3 + 2) x + (√3) 2 = 0
x² - (2 + √3) x + 2√3 = 0
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