# Solve the equation (x + 1) (x-2) = x-4 and find: a) the sum of the squares of the roots of this equation

**Solve the equation (x + 1) (x-2) = x-4 and find: a) the sum of the squares of the roots of this equation b) the difference of the cubes of the roots of this equation**

Move the right side of the equation to the left side of the equation with a minus sign.

The equation will turn from

(x – 2) (x + 1) = x – 4

in

– x + 4 + (x – 2) (x + 1) = 0

Let’s open the expression in the equation and get the quadratic equation

x ^ {2} – 2 x + 2 = 0

This is an equation of the form

a * x ^ 2 + b * x + c = 0

The quadratic equation can be solved using the discriminant.

D = b ^ 2 – 4 * a * c

Because

a = 1

b = -2

c = 2

, then

D = b ^ 2 – 4 * a * c = (-2) ^ 2 – 4 * (1) * (2) = -4

Because D <0, then the equation has no real roots, but there are complex roots.

x1 = (-b + √ (D)) / (2 * a)

x2 = (-b – √ (D)) / (2 * a)

or

x_ {1} = 1 + i

x_ {2} = 1 – i

Sum of Squares – (1 + i) ^ 2 + (1 – i) ^ 2 = 0

Difference of cubes of roots (1 + i) ^ 1/3 – (1 – i) ^ 1/3 = – 2 2/3 i / 2 +2 2/3 i / {2} √ {3}