Question

Solve for x
where x not equal to 5​

Answer 1

Step-by-step explanation:

${( \frac{2x}{x - 5} )}^{2} + \frac{2x}{x - 5} = 24$

$\frac{4 {x}^{2} }{(x - 5)(x - 5)} + \frac{2x}{x - 5} = 24$

$\frac{4 {x}^{2} + 2x(x - 5) }{(x - 5)(x - 5)} = 24$

$\frac{4 {x}^{2} + 2 {x}^{2} - 10x }{ {x}^{2} + 25 - 10x } = 24$

$\frac{6 {x}^{2} - 10x }{ {x}^{2} - 10x + 25} = 24$

$\frac{2(3 {x}^{2} - 5x) }{ {x}^{2} - 10x + 25 } = 24$

$\frac{3 {x}^{2} - 5x }{ {x}^{2} - 10x + 25} = 12$

$3 {x}^{2} - 5x = 12 {x}^{2} - 120x + 300$

$12 {x}^{2} - 3 {x}^{2} - 120x + 5x + 300 = 0$

$8 {x}^{2} - 115x + 300 = 0$

8x² -115x +300 =0

8x² -88x - 27x +300 =0

8x(x-11) -27(x-11) =0

(8x - 27)(x-11) =0

8x - 27 = 0 or x -11 =0

8x = 27 or x = 11

x = 27/8 or x = 11

NOTE:

Here the 88x and 27x are the approximate values and the values of x are approximate values.

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