(x-2)^2/3-(2x-3)(2x+3)/8+(x-4)^2/6=0

2 min read Jun 17, 2024

Solving the Equation: (x-2)^2/3 - (2x-3)(2x+3)/8 + (x-4)^2/6 = 0

This equation presents a challenge due to its fractions and the need for simplification before solving for 'x'. Let's break down the process step-by-step:

1. Find the Least Common Multiple (LCM)

The denominators of the fractions are 3, 8, and 6. The LCM of these numbers is 24.

2. Multiply the Equation by the LCM

Multiplying every term of the equation by 24 eliminates the fractions:

24 * [(x-2)^2/3] - 24 * [(2x-3)(2x+3)/8] + 24 * [(x-4)^2/6] = 0

This simplifies to:

8(x-2)^2 - 3(2x-3)(2x+3) + 4(x-4)^2 = 0

3. Expand the Squares and Products

Expand the squares and the product of the binomials:

8(x^2 - 4x + 4) - 3(4x^2 - 9) + 4(x^2 - 8x + 16) = 0

4. Simplify the Equation

Distribute and combine like terms:

8x^2 - 32x + 32 - 12x^2 + 27 + 4x^2 - 32x + 64 = 0

-16x + 123 = 0

5. Solve for 'x'

Isolate 'x':

-16x = -123

x = -123 / -16

Solution

Therefore, the solution to the equation (x-2)^2/3 - (2x-3)(2x+3)/8 + (x-4)^2/6 = 0 is:

x = 123/16