Question

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

Answer 1

Let a and b be the roots
a+b=5-----------------(1)(sum of zeroes)
a^3+b^3=35
(a+b)(a^2 +b^2 - ab) =35
5(a^2 +b^2 +2ab-3ab)=35
(a+b)^2 - 3ab=7
5^2 - 3ab=7
-3ab=7-25
-3ab=-18
ab=6(product of zeroes)
Now required equation is
x^2 - 5x+18=0

Answer 2

Answer:

x2 - 5x + 6 = 0

Step-by-step explanation:

a +β = 5

a^+β^ = 35

a^+β^ = (a +β) (a2-a β +β2)

35 = 5 (a2+β2+2aβ-3a β )

7 = { (a + β ) 2 - 3aβ }

7 = (25-3aβ)

aβ = 18/3

aβ = 6

so, x2-5x+6=0