FORMATION OF THE QUADRATIC EQUATION FROM COMPLEX ROOTS

Form quadratic equations with the following given numbers as its roots.

Problem 1 :

3 + i , 3 – i

Solution :

α  = 3 + i and β = 3 - i

α + β = 3 + i + 3 - i

α + β = 6

α β = (3 + i) (3 - i)

α β = 32 - i2

α β = 9 - (-1)

α β = 10

x2 - 6x + 10 = 0

Form quadratic equations with the following given numbers as its roots.

Problem 2 :

4 + 5i, 4 - 5i

Solution :

α  = 4 + 5i and β = 4 - 5i

α + β = 4 + 5i + 4 - 5i

α + β = 8

α β = (4 + 5i) (4 - 5i)

α β = 42 - (5i)2

= 16 -25(-1)

= 16 + 25

= 41

Problem 3 :

Find a quadratic polynomial whose zeroes are 2 + √3 and 2 – √3

Solution :

Here α = 2 + √3 and β = 2 – √3

α + β = 2 + √3 + 2 – √3 ==> 4

α β = (2 + √3)(2 – √3)

= 22 - √32

= 4 - 3

= 1

x2 - 4x + 1 = 0

So, the required polynomial is x2 - 4x + 1 = 0.

Problem 4 :

What is an equation whose roots are 5+√2 and 5−√2

Solution :

α = 5+√2, β = 5-√2

α + β = 5 + √2 + 5 - √2

α + β = 10

αβ = (5+√2) (5-√2)

αβ = 52-√22

 = 25 - 2

= 23

x2 - (α + β) x + α β = 0

x2 - 10x + 23 = 0

Problem 5 :

Find the quadratic polynomial with rational coefficients which has 1/(3+2√2) as a root.

Solution :

𝛼 = 13+22 and 𝛽 = 13-22𝛼+𝛽 = 13+22 + 13-22= 3-22 + 3+223+22(3-22)= 632-222= 69-4(2)= 69-8= 6
𝛼 = 13+22 and 𝛽 = 13-22𝛼𝛽 = 13+22 x 13-22= 13+22(3-22)= 132-222= 19-4(2)= 19-8= 1

x2 - 6x + 1 = 0

Problem 6 :

Form quadratic equations with the following given numbers as its roots.

2 + 3, 2 - 3

Solution :

α = 2 + √3, β = 2 - √3

α + β = 2 + √3 + 2 - √3

α + β = 4

αβ = (2 + √3) (2 - √3)

αβ = 22-√32

 = 4 - 3

= 1

x2 - (α + β) x + α β = 0

x2 - 4x + 1 = 0

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