Find the quadratic polynomial with zeros 2 + √3 and 2 - √3.?

The quadratic polynomial with zero a and b
= (x - a)(x - b).

The required quadratic polynomial
= [x - (2 + √3)] [x - (2 - √3)]
= x² - [(2 + √3) + (2 - √3)]x + (2 + √3)(2 - √3)
= x² - [2 + √3 + 2 - √3]x + [2² - (√3)]
= x² - 4x + 1

Let p(x)=a[x-(2+sqr(3))][x-(2-sqr(3))]
=>
p(x)=a[(x-2)-sqr(3)][(x-2)+sqr(3)]
=>
p(x)=a[(x-2)^2-3]
=>
p(x)=a(x^2-4x+1)
where a is an undetermined constant.

x = 2 ± √3

so, x - 2 = ±√3

i.e. (x - 2)² = 3

=> x² - 4x + 1 = 0 

:)>

Many polynomials satisfy those conditions, so of course it is not possible to identify "the" polynomial. Here is the whole family in x:

a[x - 2 - √(3)][x - 2 + √(3)], where a ≠ 0

(x - (a + b * sqrt(c))) * (x - (a - b * sqrt(c))) =>
(x - a - b * sqrt(c)) * (x - a + b * sqrt(c))

(m - n) * (m + n) = m^2 - n^2

(x - a)^2 - b^2 * c =>
x^2 - 2ax + a^2 - b^2 * c

(x - (2 + 1 * sqrt(3))) * (x - (2 - 1 * sqrt(3)))

a = 2 , b = 1 , c = 3

x^2 - 2 * 2 * x + 2^2 - 1^2 * 3 =>
x^2 - 4x + 4 - 3 =>
x^2 - 4x + 1


x^2 - 4x + 1 = 0
x = (4 +/- sqrt(16 - 4)) / 2
x = (4 +/- sqrt(12)) / 2
x = (4 +/- 2 * sqrt(3)) / 2
x = 2 +/- sqrt(3)

Apply the exact same reasoning already given you you in

https://ca.answers.yahoo.com/question/index?qid=20200804032416AAuI0N0

 the quadratic polynomial with zeros 2 + √3 and 2 - √3.
 roots 1 - 5 x + x^3 = 0

In general a quadratic polynomial in x  with zeros a & b is of form (x-a)(x-b).;
Here a = 2 + rt3 and b = 2 - rt3, where rt = ''square root of''.;
Then the required polynomial is (x-2+rt3)(x-2-rt3) = (x-2)^2 - (rt3)^2 = x^2-4x+4-3 = x^2-4x+1.

Many ways to skin a cat: Let's assume that a will be 1. If not, we can have a family of polynomials of a(x^2+bx+c)
We have simultaneous equations:
(-b+sqrt(b^2-4c))/2 = 2+sqrt(3)
(-b-sqrt(b^2-4c))/2=2-sqrt(3)
Let's add them together to see what happens
-2b/2=4, so b = -4
Substituting back in
(4+sqrt(16-4c))/2=2+sqrt(3)
2+sqrt(16-4c)/2 = 2+sqrt(3)
2+sqrt(4(4-c))/2=2+sqrt(3)
sqrt(4-c)=sqrt(3)
4-c=3, so c=1 and we have x^2-4x+1.
and we can have any a so a(x^2-4x+1) has those two zeroes.

You could also recognize that the polynomial would be (x-2-sqrt(3))(x-2+sqrt(3)) (multiplied by any constant factor) would do the job.
So, that's x^2+(-2-sqrt(3)-2+sqrt(3))x + (2-sqrt(3))(2+sqrt(3))
x^2-4x+2^2-sqrt(3)^2= x^2-4x+1