Question

The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35. find the equation
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Answer 1

Let x and y be the roots of the polynomial respectively.

According to the given condition,

x + y = 5

x^3 + y^3 = 35

x^3 + y^3 = (x +y)^3 - 3xy (x + y)

35 = 5^3 - 3xy (5)

35 = 125 - 15xy

15xy = 125 - 35

xy = 6

We have

sum of zeroes (x + y ) = 5

Product of zeroes (xy) = 6

Therefore,

p(x) = x^2 - (sum of zeroes)x + (product of zeroes)= 0

p(x) = x^2 - 5x + 6 = 0

Answer 2

Solution :-

Let α and β be the twoo roots of a quadratic equation

Sum of two roots = α + β = 5

Sum of their cubes = α³ + β³ = 35

⇒ α³ + β³ = (α + β)³ - 3αβ(α + β)

Since (x + y)³ = x³ + y³ + 3xy(x + y)

By sustituting the known values

⇒ 35 = 5³ - 3αβ(5)

⇒ 35 = 125 - 15αβ

⇒ 15αβ = 125 - 35

⇒ 15αβ = 90

⇒ αβ = 90/15

⇒ αβ = 6

i.e Product of zeroes = 6

Quadratic polynomial :

x² - x(α + β) + αβ = 0

By substituting the known values

⇒ x² - x(5) + 6 = 0

⇒ x² - 5x + 6 = 0

Therefore the Quadratic equation is x² - 5x + 6 = 0.