Question

write the polynomial whose zeroes are 2+√3 and 2-√3

Answer 1

The polynomial whose zeroes are $\bold{2+\sqrt{3}}$  and $\bold{2-\sqrt{3}}$ is $\bold{x^{2}-4 x+1}$

Solution:  

Since there are 2 zeroes, the polynomial is of degree 2.

Let a  and b represents the zeroes of the polynomial.

Let $a=2+\sqrt{3}$ and $b=2-\sqrt{3}$

Since, the polynomial is of degree 2

Polynomial would be of the form $\bold{x^{2}-(a+b) x+a b}$  

Therefore computing the values,

$\bold{a+b=(2+\sqrt{3})+(2-\sqrt{3})=4}$

$\bold{a b=(2+\sqrt{3}) \times(2-\sqrt{3})=4-3=1}$

Therefore, the polynomial is $\bold{x^{2}-4 x+1}$  

Answer 2

Answer:

Required quadratic polynomial x²-4x+1.

Step-by-step explanation:

$Let\:the \: quadratic \: polynomial\\be\:ax^{2}+bx+c,\\a≠0\:and\:its\: zeroes\:be\\</p><p>\alpha, \:\beta.$

$Here,\:\alpha = 2+\sqrt{3},\\\beta = 2-\sqrt{3}$

$i) Sum\:of\:the\: zeroes\\=\alpha+\beta \\= 2+\sqrt{3}+2-\sqrt{3}\\=4$

$ii) Product\:of\:the\:zeroes\\=\alpha\beta\\=(2+\sqrt{3})(2-\sqrt{3})\\=2^{2}-(\sqrt{3})^{2}\\=4-3\\=1$

Therefore the Quadratic polynomial ax²+bx+c is

$k[x^{2}-(\alpha+\beta)x+\alpha\beta],\\where\:k\:is\:a\\constant$

$=k[x^{2}-4x+1]$

We can put different values of k.

When k = 1 , the quadratic polynomial will be x²-4x+1.

•••♪