(x-5)^4-5(x-5)^2-36=0

2 min read Jun 17, 2024

Solving the Equation (x-5)^4 - 5(x-5)^2 - 36 = 0

This equation may look intimidating at first, but we can solve it by employing a clever substitution and our knowledge of quadratic equations.

Let's simplify the equation by substituting a new variable. Let y = (x-5)^2. Substituting this into our original equation, we get:

y^2 - 5y - 36 = 0

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

y = (-b ± √(b^2 - 4ac)) / 2a

Where a = 1, b = -5, and c = -36. Plugging in the values:

y = (5 ± √((-5)^2 - 4 * 1 * -36)) / 2 * 1 y = (5 ± √(169)) / 2 y = (5 ± 13) / 2

This gives us two possible solutions for y:

Now we need to substitute back our original expression for y, (x-5)^2:

Solving for x in each equation:

For (x-5)^2 = 9:

For (x-5)^2 = -4:

This equation has no real solutions since the square of a real number cannot be negative.

Therefore, the solutions to the equation (x-5)^4 - 5(x-5)^2 - 36 = 0 are x = 8 and x = 2.