Question

X+y=4 and x^3+y^3=44 then write a quadratic equation whose roots are x and y

Answer 1

${\huge {\mathfrak {\orange {Answer :-}}}}$

We have,

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

Or, 4^3 = 44 + 3xy(4)

Or, 64 - 44 = 12xy

Or, 20 = 12xy

Or, xy = 20/12 = 5/3

The quadratic equation whose roots will be x and y is

Or, a^2 + (x + y)a + xy = 0

Or, a^2 + 4a + 5/3 = 0

Or, 3a^2 + 12a + 5 = 0

$<b>➡️ The equation established is 3a^2 + 12a + 5 = 0 ⬅️</b>$

That's it..

Answer 2

Step-by-step explanation:

We have,

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

Or, 4^3 = 44 + 3xy(4)

Or, 64 - 44 = 12xy

Or, 20 = 12xy

Or, xy = 20/12 = 5/3 .