Question #385c9

1 Answer
Feb 7, 2018

Complex solutions:

x_1=(-1+i\sqrt(3))/2 and x_2=(-1-i\sqrt(3))/2

Check the explanation for the solution using sum and product

Explanation:

x_1+x_2=-1 " and " x_1 x_2=1

x_1+x_2=-1=>(x_1+x_2)^2=1=>x_1^2+x_2^2+2x_1 x_2=1
x_1^2 + x_2 ^2 + 2 \cdot 1=1=>x_1^2 + x_2^2 = -1

So, by using sum and product we proof that the solutions are non real, because if x_1 and x_2" were reals, the sum of their squares couldn't be negative.

x_1 = a+b i
x_2 = a-b i
(this came from the fact that if a+bi is a root, so a-bi is a root too, and the quadratic equation has two roots)

With a and b real, and b positive.

x_1+x_2=-1=>a=-1/2
So...
x_1 = -1/2 + bi" and "x_1=-1/2-bi

Lets use the product
x_1 x_2 = 1=>(-1/2+bi)(-1/2-bi)=1
1/4+b^2=1=>b^2=3/4=>b=\sqrt(3)/2

Then...

x_1=(-1+i\sqrt(3))/2 and x_2=(-1-i\sqrt(3))/2