Math

# Find the roots of the following by factorization​

\underline{\textsf{Given:}}

Given:

\mathsf{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{5}{6}}

x−1

x+1

x+1

x−1

=

6

5

\underline{\textsf{To find:}}

To find:

\textsf{Factors of}Factors of

\mathsf{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{5}{6}}

x−1

x+1

x+1

x−1

=

6

5

\underline{\textsf{Solution:}}

Solution:

\textsf{Consider,}Consider,

\mathsf{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{5}{6}}

x−1

x+1

x+1

x−1

=

6

5

\mathsf{\dfrac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}=\dfrac{5}{6}}

(x−1)(x+1)

(x+1)

2

−(x−1)

2

=

6

5

\mathsf{\dfrac{(x+1)^2-(x-1)^2}{x^2-1}=\dfrac{5}{6}}

x

2

−1

(x+1)

2

−(x−1)

2

=

6

5

\mathsf{\dfrac{(x^2+1+2x)-(x^2+1-2x)}{x^2-1}=\dfrac{5}{6}}

x

2

−1

(x

2

+1+2x)−(x

2

+1−2x)

=

6

5

\mathsf{\dfrac{4x}{x^2-1}=\dfrac{5}{6}}

x

2

−1

4x

=

6

5

\mathsf{6(4x)=5(x^2-1)}6(4x)=5(x

2

−1)

\mathsf{24x=5x^2-5}24x=5x

2

−5

\mathsf{5x^2-24x-5=0}5x

2

−24x−5=0

\mathsf{5x^2-25x+x-5=0}5x

2

−25x+x−5=0

\mathsf{5x(x-5)+1(x-5)=0}5x(x−5)+1(x−5)=0

\mathsf{(5x+1)(x-5)=0}(5x+1)(x−5)=0

\implies\boxed{\mathsf{x=5,\dfrac{-1}{5}}}⟹

x=5,

5

−1

\textsf{The factors are (5x+1) and (x-5)}The factors are (5x+1) and (x-5)

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hey mate its urgent please solve this it is very urgent

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