Write a quadratic equation whose roots are 5 + i radical 2 and 5 – i radical 2
____ x^2 + _____ x+ ______=0
Or
use the old sum of roots and product of roots property
sum of roots = 5+i√2 + 5-i√2 = 10
product of roots = (5+i√2)(5-i√2)
=25-2i^2 = 25+2 = 27
x^2 - (sum)x + product = 0
x^2 - 10x + 27 = 0
I've also preferred to do the complex multiply stuff using the fact that
(a-b)(a+b) = a^2-b^2, so
(a-bi)(a+bi) = a^2+b^2
(x-(5-√2i))(x-(5+√2i))
((x-5)+√2i)((x-5)-√2i)
(x-5)^2 + √2^2
x^2-10x+25 + 2
x^2-10x+27
To write a quadratic equation with the given roots, we have to use the fact that if alpha and beta are the roots of a quadratic equation, then the equation can be expressed as follows:
(x - alpha) * (x - beta) = 0.
In this case, the given roots are 5 + i√2 and 5 - i√2. To form the equation, we substitute these values into the equation as follows:
(x - (5 + i√2)) * (x - (5 - i√2)) = 0.
Now, let's simplify this equation:
(x - 5 - i√2) * (x - 5 + i√2) = 0
(x - 5)^2 - (i√2)^2 = 0
(x - 5)^2 - [i^2 * (√2)^2] = 0
(x - 5)^2 - [(-1) * 2] = 0
(x - 5)^2 + 2 = 0
Finally, we have the quadratic equation with the given roots:
x^2 - 10x + 27 = 0.
(x - x1) (x - x2) = 0
(x - 5 - isqrt2) (x -5 +isqrt2) =0
distributive property multiply
x (x -5 +isqrt2) = x^2 -5 x + ix sqrt 2
-5 (x -5 +isqrt2)= -5 x +25 -5isqrt 2
-isqrt2(x -5 +isqrt2)=-xisqrt2+5isqrt2 +2
add
= x^2 -10 x +27