Question

Please give answer the question

Answer 1

solve it further to get answer

Answer 2

$\dfrac{(2x + 1)^2 + (2x - 1)^2 }{(2x + 1)^2 - (2x - 1)^2} = \dfrac{17}{8}$

Cross Multiply:

8 [ (2x + 1)² + (2x - 1)² ] = 17 [ (2x + 1)² - (2x -1)²]

Expand with (a ± b)² = a² ± 2ab + b²:

8 [ (4x² + 4x + 1) + (4x² - 4x +  1) ] = 17 [ (4x² + 4x + 1)  -  (4x² - 4x +  1) ]

Remove brackets:

8 [ 4x² + 4x + 1 + 4x² - 4x +  1 ] = 17 [ 4x² + 4x + 1 -  4x² + 4x -  1 ]

Combine like terms:

8 [ 8x² +  2 ] = 17 [ (8x]

Distribute 8 and 17:

64x² + 16 = 136x

Move all terms to the LHS:

64x² - 136x + 16 = 0

Divide by 8 through:

8x² - 17x +2 = 0

Factorise:

(x - 2)(8x - 1) = 0

Apply zero product property:

x = 2 or 8x = 1

x = 2 or x = 1/8

Answer: x = 2 or x = 1/8