(x-6)^4+(x-8)^4=16

3 min read Jun 17, 2024

Solving the Equation (x-6)^4 + (x-8)^4 = 16

This equation might look intimidating at first glance, but we can solve it by employing some clever algebraic manipulations and a bit of ingenuity. Here's a step-by-step breakdown:

1. Recognizing the Symmetry

Notice that the equation is symmetrical about the value x = 7. This is because the terms (x-6) and (x-8) are equidistant from x = 7. This symmetry will be helpful later.

2. Substitution

Let's make a substitution to simplify the equation. Let:

y = x - 7

This means:

x - 6 = y + 1
x - 8 = y - 1

Substituting these into our original equation, we get:

(y + 1)^4 + (y - 1)^4 = 16

3. Expanding and Simplifying

Expand the powers using the binomial theorem or by repeated multiplication:

(y + 1)^4 = y^4 + 4y^3 + 6y^2 + 4y + 1
(y - 1)^4 = y^4 - 4y^3 + 6y^2 - 4y + 1

Substituting these back into the equation:

(y^4 + 4y^3 + 6y^2 + 4y + 1) + (y^4 - 4y^3 + 6y^2 - 4y + 1) = 16

Simplifying:

2y^4 + 12y^2 - 14 = 0

4. Solving the Quadratic

Notice that this equation is now quadratic in y^2. Let's substitute z = y^2:

2z^2 + 12z - 14 = 0

Divide both sides by 2:

z^2 + 6z - 7 = 0

This equation can be factored:

(z + 7)(z - 1) = 0

Therefore, z = -7 or z = 1.

5. Back Substitution

Recall that z = y^2. Substitute back to find y:

If z = -7, then y^2 = -7, which has no real solutions.
If z = 1, then y^2 = 1, which gives us y = 1 or y = -1.

6. Finding x

Finally, substitute back for x using y = x - 7:

If y = 1, then x - 7 = 1, so x = 8
If y = -1, then x - 7 = -1, so x = 6

Solution

Therefore, the solutions to the equation (x-6)^4 + (x-8)^4 = 16 are x = 6 and x = 8.