Question

x/x+1 + x+1/x =17/4​
pls answer this question.

Answer 1

Answer:

Given equation,

$\tt \implies \dfrac{x}{x + 1} + \dfrac{x + 1}{x} = \dfrac{17}{4}$

We have to find out the values of x. Here comes the solution.

$\tt \implies \dfrac{ {x}^{2} + {(x + 1)}^{2} }{x(x + 1)} = \dfrac{17}{4}$

$\tt \implies \dfrac{ {x}^{2} + {x}^{2} + 2x + 1}{x(x + 1)} = \dfrac{17}{4}$

$\tt \implies \dfrac{2{x}^{2} + 2x + 1}{{x}^{2} + x} = \dfrac{17}{4}$

On transposing, we get,

$\tt \implies 4(2{x}^{2} + 2x + 1)= 17( {x}^{2} + x)$

$\tt \implies8{x}^{2} + 8x + 4= 17{x}^{2} + 17x$

$\tt \implies 17{x}^{2} - 8 {x}^{2} + 17x - 8x - 4 = 0$

$\tt \implies 9{x}^{2} + 9x - 4 = 0$

$\tt \implies 9{x}^{2} - 3x + 12x - 4 = 0$

$\tt \implies 3x(3x - 1) + 4(3x -1)= 0$

$\tt \implies (3x + 4)(3x -1)= 0$

$\tt \implies (3x + 4) = 0 \: or \: (3x -1)= 0$

$\tt \implies x= \dfrac{ - 4}{3} , \dfrac{1}{3}$

So, the values of x are - 4/3 and 1/3.

Answer:

• x = -4/3, 1/3.

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