(x2-x)2+5(x2-x)+4=0

3 min read Jun 17, 2024

Solving the Quadratic Equation: (x^2-x)^2 + 5(x^2-x) + 4 = 0

This equation might look intimidating at first glance, but we can solve it using a clever substitution and the familiar quadratic formula.

Let's start by making a substitution to simplify the equation. Let y = x^2 - x. Now we can rewrite the equation as:

y^2 + 5y + 4 = 0

This is a much more manageable quadratic equation.

We can now solve for y using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

Where:

Substituting the values into the formula, we get:

y = (-5 ± √(5^2 - 4 * 1 * 4)) / 2 * 1

y = (-5 ± √9) / 2

This gives us two possible solutions for y:

y1 = -1

y2 = -4

Now we need to substitute back our original expression for y and solve for x:

For y1 = -1:

This quadratic equation doesn't factor easily. We can use the quadratic formula again to find the solutions for x:

x = (1 ± √(1^2 - 4 * 1 * 1)) / 2 * 1

x = (1 ± √(-3)) / 2

x = (1 ± i√3) / 2

Where i is the imaginary unit (√-1). This gives us two complex solutions for x.

For y2 = -4:

Again, this equation doesn't factor easily. Using the quadratic formula:

x = (1 ± √(1^2 - 4 * 1 * 4)) / 2 * 1

x = (1 ± √(-15)) / 2

x = (1 ± i√15) / 2

This also gives us two complex solutions for x.

In conclusion, the equation (x^2-x)^2 + 5(x^2-x) + 4 = 0 has four complex solutions: