Solve: y=x^2 and y^2=27x.

Answer 1

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 )