# HOW TO FIND THE QUADRATIC POLYNOMIAL WHOSE ZEROES ARE GIVEN

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.

## Finding Quadratic Polynomial When Zeroes are Given

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 = 0α + 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 = 0α + 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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