find the roots of the following by factorization
Answers
Step-by-step explanation:
\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
\underline{\textsf{Answer:}}
Answer:
\textsf{The factors are (5x+1) and (x-5)}The factors are (5x+1) and (x-5)
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