Solve: y=x^2 and y^2=27x.
Replace
y with x²
in
y² = 27 x
( x² )² = 27 x
x⁴ = 27 x
Subtract 27 x to both sides
x⁴ - 27 x = 0
x ( x³ - 27 ) = 0
x ( x³ - 3³ ) = 0
For ( x³ - 3³ ) apply the difference of cubes formula:
a³ - b³ = ( a - b ) ( a² + a b + b² )
with a = x and b = 3
x³ - 3³ = ( x - 3 ) ( x² + x ∙ 3 + 3² )
x³ - 3³ = ( x - 3 ) ( x² + 3 x + 9 )
So
x ( x³ - 3³ )
become
x ( x - 3 ) ( x² + 3 x + 9 )
Now you must solve:
x ( x³ - 3³ ) = 0
x ( x - 3 ) ( x² + 3 x + 9 ) = 0
The equation will be equal to zero when the terms of the equation are equal to zero.
This means you need to solve:
x = 0 , x - 3 = 0 , x² + 3 x + 9 = 0
1 condition:
x = 0
2 condition:
x - 3 = 0
Add 3 to both sides
x = 3
3 condition:
x² + 3 x + 9 = 0
The solutions are:
x = - 3 / 2 + i ∙ 3 √3 / 2
and
x = - 3 / 2 - i ∙ 3 √3 / 2
Now replace this value in equation:
y = x²
1.
x = 0
y = 0² = 0
2.
x = 3
y = 3² = 9
3.
x = - 3 / 2 + i ∙ 3 √3 / 2
y = ( - 3 / 2 + i ∙ 3 √3 / 2 )² = - 9 / 2 - i ∙ 9√3 / 2 = - 9 / 2 ( 1 + i √3 )
4.
x = - 3 / 2 - i ∙ 3 √3 / 2
y = ( - 3 / 2 - i ∙ 3 √3 / 2 )² = - 9 / 2 + i ∙ 9√3 / 2 = - 9 / 2 ( 1 - i √3 )
This equation have 4 solutions:
x = 0 , y = 0
x = 3 , y = 9
x = - 3 / 2 + i ∙ 3 √3 / 2 , y = - 9 / 2 ( 1 + i √3 )
x = - 3 / 2 - i ∙ 3 √3 / 2 , y = - 9 / 2 ( 1 - i √3 )