How to obtain a quadratic equation, with integer coefficient, having roots 2+i√5 and 2-i√5 ?

Question

Ariana Alvarez · Accepted Answer

#y=x^2-4x+9# #"given the roots of a polynomial say "x=a" and "x=b# #"then the factors of the polynomial are " # #(x-a)" and " (x-b)# #"the polynomial is then the product of the factors"# #p(x)=(x-a)(x-b)# #"here the roots are "x=2+isqrt5" and "x=2-isqrt5# #rArr"factors are "(x-2+isqrt5)" and "(x-2-isqrt5)# #rArry=((x-2)+isqrt5)((x-2)-isqrt5)# #color(white)(rArry)=(x-2)^2-5i^2# #color(white)(rArry)=x^2-4x+4+5# #color(white)(rArry)=x^2-4x+9#

Xavier Alonso · Answer

undefined To obtain a quadratic equation with integer coefficients having roots $2 + i\sqrt{5}$ and $2 - i\sqrt{5}$, you can use the fact that complex roots occur in conjugate pairs for polynomials with real coefficients.

Given the roots $r_1 = 2 + i\sqrt{5}$ and $r_2 = 2 - i\sqrt{5}$, the quadratic equation can be expressed as:

$$ (x - r_1)(x - r_2) = 0 $$

Expanding this expression, we get:

$$ (x - (2 + i\sqrt{5}))(x - (2 - i\sqrt{5})) = 0 $$
$$ (x - 2 - i\sqrt{5})(x - 2 + i\sqrt{5}) = 0 $$

Now, using the difference of squares, we can simplify this expression:

$$ (x - 2)^2 - (i\sqrt{5})^2 = 0 $$
$$ (x - 2)^2 - (-5) = 0 $$
$$ (x - 2)^2 + 5 = 0 $$

Expanding further:

$$ x^2 - 4x + 4 + 5 = 0 $$
$$ x^2 - 4x + 9 = 0 $$

So, the quadratic equation with integer coefficients having roots $2 + i\sqrt{5}$ and $2 - i\sqrt{5}$ is $x^2 - 4x + 9 = 0$.