How to Find the Quadratic Polynomial Whose Zeros are Given ?
The general form of any quadratic equation will be
x2 - (α + β) x + α β = 0
Here α and β are the zeroes of polynomial.
Question 1 :
Find the quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 ,-1
Solution :
α = 1/4
β = -1
General form of a quadratic equation
x² – (α + β) x + α β = 0
α + β = (1/4) + 1 = (1 + 4)/4 = 5/4 |
α β = (1/4) ( 1) = 1/4 |
x² – (5/4) x + (1/4) = 0
4 x² – 5 x + 1 = 0
(ii) √2, 1/3
Solution :
α = √2
β = 1/3
General form of a quadratic equation
x² – (α + β) x + α β = 0
α + β = √2 + (1/3) = (3√2 + 1)/3 |
α β = (√2) (1/3) = √2/3 |
x² – [(3√2 + 1)/3] x + (√2/3) = 0
3 x² – (3√2 + 1) x + √2 = 0
(iii) 0, √5
Solution :
α = 0
β = √5
General form of a quadratic equation
x² – (α + β) x + α β = 0 α + β = 0 + √5 = √5 |
α β = (0) (√5) = 0 |
x²– √5 x + (0) = 0
x² –√5 x = 0
(iv) 1, 1
Solution :
α = 1
β = 1
General form of a quadratic equation
x² – (α + β) x + α β = 0
α + β = 1 + 1 = 2 |
α β = (1) (1) = 1 |
x² – 2 x + 1 = 0
(v) -1/4, 1/4
Solution :
α = -1/4
β = 1/4
General form of a quadratic equation
x² – (α + β) x + α β = 0 α + β = -1/4 + 1/4 = 0 |
α β = (-1/4) (1/4) = -1/16 |
x² – 0 x + (-1/16) = 0
16 x² – 1 = 0
(vi) 4, 1
Solution :
α = 4
β = 1
General form of a quadratic equation
x² – (α + β) x + α β = 0
α + β = 4 + 1 = 5 |
α β = 4 (1) = 4 |
x² – 5 x + 4 = 0
After having gone through the stuff given above, we hope that the students would have understood, how to find the quadratic polynomial whose zeros are given.
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