(x^2-2x-5)^2-2(x^2-2x-3)-4=0

Jasonbradley

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Jun 17, 2024

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Solving the Equation (x^2 - 2x - 5)^2 - 2(x^2 - 2x - 3) - 4 = 0

This equation looks complex, but we can solve it using a few techniques. Here's how:

1. Substitution

Let's simplify the equation by substituting a variable for the repeated expression. Let:

y = x^2 - 2x - 5

Now our equation becomes:

y^2 - 2(y + 2) - 4 = 0

2. Expanding and Simplifying

Expand the equation:

y^2 - 2y - 4 - 4 = 0

Combine like terms:

y^2 - 2y - 8 = 0

3. Factoring

Now we have a quadratic equation in terms of y. We can factor this equation:

(y - 4)(y + 2) = 0

This gives us two possible solutions for y:

• y = 4
• y = -2

4. Back Substitution

Now we need to substitute back the original expression for y:

• x^2 - 2x - 5 = 4
• x^2 - 2x - 5 = -2

5. Solving the Quadratic Equations

Let's solve each of these quadratic equations:

For x^2 - 2x - 5 = 4

• x^2 - 2x - 9 = 0

We can use the quadratic formula to solve for x:

• x = (2 ± √(2^2 - 4 * 1 * -9)) / (2 * 1)
• x = (2 ± √40) / 2
• x = (2 ± 2√10) / 2
• x = 1 ± √10

For x^2 - 2x - 5 = -2

• x^2 - 2x - 3 = 0

We can factor this equation directly:

• (x - 3)(x + 1) = 0
• x = 3
• x = -1

6. Final Solutions

Therefore, the solutions for the original equation (x^2 - 2x - 5)^2 - 2(x^2 - 2x - 3) - 4 = 0 are:

• x = 1 + √10
• x = 1 - √10
• x = 3
• x = -1