Question

$\frac{1 + x + {x}^{2} }{1 - x + {x}^{2} } = \frac{13(1 + x)}{14(1 - x)}$
Find the value of x..

$(a) \frac{1}{3}$
$(b)3$
$(c) \frac{2}{3}$
$(d) \frac{3}{2}$
​

Answer 1

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$\frac{1 + x + {x}^{2} }{1 - x + {x}^{2} } = \frac{13(1 + x)}{14(1 - x)}$

Find the value of x..

$(a) \frac{1}{3}$

$(b)3$

$(c) \frac{2}{3}$

$(d) \frac{3}{2}$

$\huge{\mathcal{\underline\red{Answer}}}$

$⇒ \frac{1 + x + {x}^{2} }{1 - x + {x}^{2} } = \frac{13(1 + x)}{14(1 - x)}$

Do cross multiply

$⇒(1 + x + {x}^{2} ) \times 14(1 - x) = (1 - x + {x}^{2} ) \times 13(1 + x)$

$⇒14(1 - {x}^{3} ) = 13(1 + {x}^{3})$

$⇒14 - 14 {x}^{3} = 13 + 13 {x}^{3}$

$⇒1 = 27 {x}^{3}$

$⇒ \frac{1}{27} = {x}^{3}$

$⇒( \frac{1}{3} )^{3} = {x}^{3}$

$⇒x = \frac{1}{3}$

Answer 2

Answer:

⇒

1−x+x

2

1+x+x

2

=

14(1−x)

13(1+x)

Do cross multiply

⇒(1 + x + {x}^{2} ) \times 14(1 - x) = (1 - x + {x}^{2} ) \times 13(1 + x)⇒(1+x+x

2

)×14(1−x)=(1−x+x

2

)×13(1+x)

⇒14(1 - {x}^{3} ) = 13(1 + {x}^{3}) ⇒14(1−x

3

)=13(1+x

3

)

⇒14 - 14 {x}^{3} = 13 + 13 {x}^{3} ⇒14−14x

3

=13+13x

3

⇒1 = 27 {x}^{3} ⇒1=27x

3

⇒ \frac{1}{27} = {x}^{3} ⇒

27

1

=x

3

⇒( \frac{1}{3} )^{3} = {x}^{3} ⇒(

3

1

)

3

=x

3

⇒x = \frac{1}{3} ⇒x=

3

1