Question

pls find the roots of the following equation.
pls send the photo of the solution.​

Answer 1

Step-by-step explanation:

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Answer 2

2 and -5

Step-by-step explanation:

Given:

$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$

To find:

The roots of equation.

What you have to do ?

1. Take L.CM

2. Subtract or Add

3. Simplify and get answer...

Solution:

$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3 \\ \\ [\textsf{By Taking L.C.M}] \\ \\ \frac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3 \\ \\ \frac{( {x}^{2} + x + 2x + 2) + ( {x}^{2} - x - 2x + 2)}{ {x}^{2} - x + 2x - 2 } = 3 \\ \\ \frac{ {x}^{2} + {x}^{2} + 3x - 3x + 2 + 2}{ {x}^{2} + x - 2 } = 3 \\ \\ 2 {x}^{2} + 4= 3( {x}^{2} + x - 2) \\ \\ [\textsf{By Cross Multiply}]\\ \\ 2 {x}^{2} + 4= 3 {x}^{2} + 3x - 6 \\ \\ 3 {x}^{2} - 2 {x}^{2} + 3x - 6 - 4= 0 \\ \\ {x}^{2} + 3x - 10 = 0 \\ \\ {x}^{2} + 5x - 2x - 10 = 0 \\ \\ x(x + 5) - 2(x + 5) = 0 \\ \\ (x + 5)(x - 2) = 0$

Either, x + 5 = 0 ⟹ x = - 5

Or, x - 2 = 0 ⟹ x = 2

Therefore, -5 and 2 are the roots of given equation.