Question

Find the quadratic polynomial, the sum of whose zeros is \sqrt {2} and their product is -12. Hence, find the zeros of the polynomial.

Answer 1

let a and b are two roots

we are given a+b=sqrt(2)
and 
ab=-12

thus (a+b)^2= a^2+b^2+2ab
hence
2=a^2+b^2+2(-12)
so,26=a^2+b^2.............(1)
also (a-b)^2=a^2+b^2-2ab
                 = 26 - 2(-12)
                  =50
(a-b)=sqrt(50)...........(2)

solving (1) and (2) 

a= (5sqrt(2)+sqrt(2))/2
b= (-5sqrt(2)+sqrt(2))/2

Answer 2

Let α and β be the zeros of the equation:
Sum of the zeros α+β = -b/a =√2...........(!)
Product of zeros αβ = c/a = -12..................(!!)
from (!!) α = -12/β .....................................(!!!)
Substitute (!!!) in (!)
-12/β + β = √2
(-12β+β²)/β = √2
-12β+β²=√2β   -12+β=√2
β²=√2β+12β
β² = (√2+12)β
β = √2+12

β in (!!!)
α = -12 / √2+12