(2x^2-3x-1)^2-3(2x^2-3x-5)-16=0

3 min read Jun 16, 2024

Solving the Equation: (2x^2-3x-1)^2-3(2x^2-3x-5)-16=0

This equation appears complex, but we can simplify it by using substitution and then applying the quadratic formula. Here's a step-by-step solution:

1. Substitution:

Let's make the equation easier to handle by substituting y = 2x^2 - 3x - 1.

This transforms our equation into:

y^2 - 3(y - 4) - 16 = 0

2. Simplifying the Equation:

Expanding and rearranging the terms, we get:

y^2 - 3y + 12 - 16 = 0

y^2 - 3y - 4 = 0

3. Solving the Quadratic Equation:

Now we have a simple quadratic equation. We can solve it using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

Where:

Plugging in these values:

y = (3 ± √((-3)^2 - 4 * 1 * -4)) / (2 * 1)

y = (3 ± √(25)) / 2

y = (3 ± 5) / 2

This gives us two possible solutions for y:

4. Back-substitution:

Now, let's substitute back the original expression for y:

For y1 = 4:

2x^2 - 3x - 1 = 4

2x^2 - 3x - 5 = 0

For y2 = -1:

2x^2 - 3x - 1 = -1

2x^2 - 3x = 0

5. Solving for x:

We now have two separate quadratic equations to solve for x. We can use the quadratic formula again or factor them directly:

For 2x^2 - 3x - 5 = 0:

For 2x^2 - 3x = 0:

6. Final Solutions:

Therefore, the solutions to the original equation (2x^2-3x-1)^2-3(2x^2-3x-5)-16=0 are: