Question

How do you find the quadratic equation, in standard form, that has the roots x= (2±√5)/3

Answer 1

if the roots are a and b, the function is

(x-a)(x-b) = 0

So, plug in
a=(2+√5)/3
b=(2-√5)/3

Answer 2

several ways:

if x = (2 ± √5)/3
then the factors are

(x - (2 +√5)/3) and (x - (2 - √5)/3)
then
(x - (2 +√5)/3)(x - (2 - √5)/3) = 0
( x - 2/3 - √5/3)(x - 2/3 + √5/3) = 0
x^2 - 2x/3 + x(√5/3) - 2x/3 + 4/9 - 2√5/9 - x√5/3 + 2√5/9 - 5/9 = 0
x^2 - 4x/3 - 1/9 = 0
times 9

9x^2 - 12x - 1 = 0

or, you can work backwards
x = (2 ± √5)/3
3x = 2 ± √5
3x - 2 = √5 OR 3x-2 = -√5
square both sides
9x^2 - 12x + 4 = 5 , both (√5)^2 and (-√5)^2 = 5
9x^2 - 12x -1 = 0


easiest way, based on the fact that for
ax^2 + bx + c = 0
the sum of the roots = -b/a, product or roots = c/a

sum of roots = (2 + √5)/3) + (2 - √5)/3 = 4/3
product of roots = (2 + √5)/3)(2 - √5)/3
= (4 - 5)/9 = -1/9
then x^2 - (4/3)x - 1/9 = 0
times 9
9x^2 - 12x - 1 = 0