The roots of the equation xx+1+x+1x=2415, x≠0,-1 are [[1]] and [[2]].

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The roots of the equation $\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0, - 1$ are [[1]] and [[2]].

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answer is - 5 2, 3 2.

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Detailed Solution

The roots of the equation

\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0, - 1

are

- \frac{5}{2}

and

\frac{3}{2}

.
Given equation is,

\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}

Then,

\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}

⟹ \frac{x^{2} + {(x + 1)}^{2}}{x (x + 1)} = \frac{34}{15}

⟹ \frac{x^{2} + x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15}

⟹ \frac{2 x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15}

⟹ 15 (2 x^{2} + 2 x + 1) = 34 (x^{2} + x)

⟹ 30 x^{2} + 30 x + 15 = 34 x^{2} + 34 x

⟹ 34 x^{2} - 30 x^{2} + 34 x - 30 x - 15 = 0

⟹ 4 x^{2} + 4 x - 15 = 0

We know that,
Discriminant for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

D = b^{2} - 4 ac

.
On comparing the equation

4 x^{2} + 4 x - 15 = 0

with standard form we get,

a = 4, b = 4, c = - 15

.
Then,

D = b^{2} - 4 ac

⟹ D = {(4)}^{2} - 4 \times 4 \times - 15

⟹ D = 16 + 240

⟹ D = 256 > 0

Discriminant is greater than zero which means the roots of the equation exist.
Since, quadratic formula for the quadratic equation

a x^{2} + bx + c = 0, a \neq 0

where a, b and c are real numbers is given by,

x = \frac{- b \pm \sqrt{D}}{2 a}

Then,

x = \frac{- 4 \pm \sqrt{256}}{2 \times 8}

⟹ x = \frac{- 4 \pm 16}{8}

⟹ x = \frac{- 1 \pm 4}{2}

On taking positive sign,

⟹ x = \frac{- 1 + 4}{2}

⟹ x = \frac{3}{2}

On taking negative sign,

⟹ x = \frac{- 1 - 4}{2}

⟹ x = \frac{- 5}{2}

Therefore, the roots of the equation

\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}

are

\frac{3}{2}

and

\frac{- 5}{2}

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