challanging-question-if-1-1-1-x-x-2-x-1-1-x-x-2-1-find-the-value-of-x-x-2-1-

Question Number 182852 by mnjuly1970 last updated on 16/Dec/22

c h a l l a n g i n g q u e s t i o n . .

i f, \frac{1}{1 + \sqrt{1 - x + x^{2}}} - \frac{x}{1 + \sqrt{1 + x + x^{2}}} = 1

f i n d t h e v a l u e o f : \frac{x}{x^{2} + 1}

Commented by Frix last updated on 15/Dec/22

The solutions of the equation are x=((6+(√(−34+10(√(13)))))/(10))±((2+(√(34+10(√(13)))))/(10))i ⇒ (x/(x^2 +1))=((3±i)/4)

The solutions of the equation are

x = \frac{6 + \sqrt{- 34 + 10 \sqrt{13}}}{10} \pm \frac{2 + \sqrt{34 + 10 \sqrt{13}}}{10} i

\Rightarrow \frac{x}{x^{2} + 1} = \frac{3 \pm i}{4}

Commented by Frix last updated on 16/Dec/22

Now you changed it. Now x=−((1+(√7)+(√(−1+2(√7))))/3) ⇒ (x/(x^2 +1))=((1−(√7))/4)

Now you changed it .

Now x = - \frac{1 + \sqrt{7} + \sqrt{- 1 + 2 \sqrt{7}}}{3} \Rightarrow

\frac{x}{x^{2} + 1} = \frac{1 - \sqrt{7}}{4}

Answered by Rasheed.Sindhi last updated on 16/Dec/22

(1/( 1+(√(1−x+x^( 2) )))) +(x/(1+(√(1+x+x^( 2) ))))=1 (x/(x^( 2) +1))=? (1/( 1+(√(x(x+(1/x)−1))))) +(x/(1+(√(x(x+(1/x)+1)))))=1 (1/( 1+(√x) (√(x+(1/x)−1))))+(x/( 1+(√x) (√(x+(1/x)+1)) ))=1 Let x+(1/x)+1=y^2 x+(1/x)−1=y^2 −2 (1/( 1+(√x) (√(y^2 −2))))+(x/( 1+y(√x) ))=1 1−(1/( 1+(√x) (√(y^2 −2))))=(x/( 1+y(√x) )) ((1+(√x) (√(y^2 −2)) −1)/(1+(√x) (√(y^2 −2))))=(x/( 1+y(√x) )) (((√x) (√(y^2 −2)) )/(1+(√x) (√(y^2 −2))))=(x/( 1+y(√x) )) ((1+(√x) (√(y^2 −2)))/( (√x) (√(y^2 −2)) ))=((1+y(√x) )/x) (1/( (√x) (√(y^2 −2)) ))+1=(1/x)+(y/( (√x))) (1/( (√x) (√(y^2 −2)) ))−(y/( (√x)))=(1/x)−1=((1−x)/x) (1/( (√(y^2 −2))))−y=((1−x)/( (√x))) ((1−y(√(y^2 −2)) )/( (√(y^2 −2))))=((1−x)/( (√x))) (((1−y(√(y^2 −2)) )/( (√(y^2 −2)))))^2 =(((1−x)/( (√x))))^2 ((1+y^2 (y^2 −2)−2y(√(y^2 −2)))/(y^2 −2))=((1+x^2 −2x)/x) ((1+y^2 (y^2 −2)−2y(√(y^2 −2)))/(y^2 −2))=x+(1/x)−2 ((1+y^2 (y^2 −2)−2y(√(y^2 −2)))/(y^2 −2))=x+(1/x)+1−3 ((1+y^2 (y^2 −2)−2y(√(y^2 −2)))/(y^2 −2))=y^2 −3 1+y^4 −2y^2 −2y(√(y^2 −2)) =(y^2 −3)(y^2 −2) 1+y^4 −2y^2 −2y(√(y^2 −2)) =y^4 −5y^2 +6 2y(√(y^2 −2)) =3y^2 −5 4y^2 (y^2 −2)=9y^4 −30y^2 +25 5y^4 −22y^2 +25=0 y^2 =((22±(√(484−500)))/(10))=((22±4i)/(10))=((11±2i)/5) ▶(x/(x^( 2) +1))=(1/(x+(1/x)))=(1/(x+(1/x)+1−1))=(1/(y^2 −1)) =(1/(((11±2i)/5)−1))=(1/((6±2i)/5))=(5/(6±2i))∙((6∓2i)/(6∓2i)) =((5(6∓2i))/(36+4))=((6∓2i)/8)=((3±i)/4)

\frac{1}{1 + \sqrt{1 - x + x^{2}}} + \frac{x}{1 + \sqrt{1 + x + x^{2}}} = 1

\frac{x}{x^{2} + 1} = ?

\frac{1}{1 + \sqrt{x (x + \frac{1}{x} - 1)}} + \frac{x}{1 + \sqrt{x (x + \frac{1}{x} + 1)}} = 1

\frac{1}{1 + \sqrt{x} \sqrt{x + \frac{1}{x} - 1}} + \frac{x}{1 + \sqrt{x} \sqrt{x + \frac{1}{x} + 1}} = 1

L e t x + \frac{1}{x} + 1 = y^{2}

x + \frac{1}{x} - 1 = y^{2} - 2

\frac{1}{1 + \sqrt{x} \sqrt{y^{2} - 2}} + \frac{x}{1 + y \sqrt{x}} = 1

1 - \frac{1}{1 + \sqrt{x} \sqrt{y^{2} - 2}} = \frac{x}{1 + y \sqrt{x}}

\frac{1 + \sqrt{x} \sqrt{y^{2} - 2} - 1}{1 + \sqrt{x} \sqrt{y^{2} - 2}} = \frac{x}{1 + y \sqrt{x}}

\frac{\sqrt{x} \sqrt{y^{2} - 2}}{1 + \sqrt{x} \sqrt{y^{2} - 2}} = \frac{x}{1 + y \sqrt{x}}

\frac{1 + \sqrt{x} \sqrt{y^{2} - 2}}{\sqrt{x} \sqrt{y^{2} - 2}} = \frac{1 + y \sqrt{x}}{x}

\frac{1}{\sqrt{x} \sqrt{y^{2} - 2}} + 1 = \frac{1}{x} + \frac{y}{\sqrt{x}}

\frac{1}{\sqrt{x} \sqrt{y^{2} - 2}} - \frac{y}{\sqrt{x}} = \frac{1}{x} - 1 = \frac{1 - x}{x}

\frac{1}{\sqrt{y^{2} - 2}} - y = \frac{1 - x}{\sqrt{x}}

\frac{1 - y \sqrt{y^{2} - 2}}{\sqrt{y^{2} - 2}} = \frac{1 - x}{\sqrt{x}}

{(\frac{1 - y \sqrt{y^{2} - 2}}{\sqrt{y^{2} - 2}})}^{2} = {(\frac{1 - x}{\sqrt{x}})}^{2}

\frac{1 + y^{2} (y^{2} - 2) - 2 y \sqrt{y^{2} - 2}}{y^{2} - 2} = \frac{1 + x^{2} - 2 x}{x}

\frac{1 + y^{2} (y^{2} - 2) - 2 y \sqrt{y^{2} - 2}}{y^{2} - 2} = x + \frac{1}{x} - 2

\frac{1 + y^{2} (y^{2} - 2) - 2 y \sqrt{y^{2} - 2}}{y^{2} - 2} = x + \frac{1}{x} + 1 - 3

\frac{1 + y^{2} (y^{2} - 2) - 2 y \sqrt{y^{2} - 2}}{y^{2} - 2} = y^{2} - 3

1 + y^{4} - 2 y^{2} - 2 y \sqrt{y^{2} - 2} = (y^{2} - 3) (y^{2} - 2)

1 + y^{4} - 2 y^{2} - 2 y \sqrt{y^{2} - 2} = y^{4} - 5 y^{2} + 6

2 y \sqrt{y^{2} - 2} = 3 y^{2} - 5

4 y^{2} (y^{2} - 2) = 9 y^{4} - 30 y^{2} + 25

5 y^{4} - 22 y^{2} + 25 = 0

y^{2} = \frac{22 \pm \sqrt{484 - 500}}{10} = \frac{22 \pm 4 \boldsymbol i}{10} = \frac{11 \pm 2 \boldsymbol i}{5}

▸ \frac{x}{x^{2} + 1} = \frac{1}{x + \frac{1}{x}} = \frac{1}{x + \frac{1}{x} + 1 - 1} = \frac{1}{y^{2} - 1}

= \frac{1}{\frac{11 \pm 2 \boldsymbol i}{5} - 1} = \frac{1}{\frac{6 \pm 2 i}{5}} = \frac{5}{6 \pm 2 i} \cdot \frac{6 \mp 2 i}{6 \mp 2 i}

= \frac{5 (6 \mp 2 i)}{36 + 4} = \frac{6 \mp 2 i}{8} = \frac{3 \pm i}{4}

Commented by manxsol last updated on 16/Dec/22

I am looking for a simpler development(desarrollo) .Greetings

I a m l o o k i n g f o r a

s i m p l e r

d e v e l o p m e n t (d e s a r r o l l o)

. G r e e t i n g s

Commented by manxsol last updated on 16/Dec/22

error in line 2y(√(y^2 −2))=3y^2 −5 4y^2 (y^2 −2)=3y^2 −5 4y^2 (y^2 −2)=9y^4 −30y^2 +25 5y^4 −22y^2 +25=0 y^2 = ((22±(√(22^2 −500)))/(10)) y^2 =((22±4i)/(10)) y^2 =((11±2i)/5) (x/(x^2 +1))=?⇒(1/(x+(1/x)))=(1/(y^2 −1)) (1/(((11±2i)/5)−1))=(1/((6±2i)/5))=(5/(6±2i))=((5(6−2i))/(40)) (x/(x^2 +1))=((3±i)/4)

e r r o r i n l i n e

2 y \sqrt{y^{2} - 2} = 3 y^{2} - 5

4 y^{2} (y^{2} - 2) = 3 y^{2} - 5

4 y^{2} (y^{2} - 2) = 9 y^{4} - 30 y^{2} + 25

5 y^{4} - 22 y^{2} + 25 = 0

y^{2} = \frac{22 \pm \sqrt{22^{2} - 500}}{10}

y^{2} = \frac{22 \pm 4 \boldsymbol i}{10}

y^{2} = \frac{11 \pm 2 \boldsymbol i}{5}

\frac{x}{x^{2} + 1} = ? \Rightarrow \frac{1}{x + \frac{1}{x}} = \frac{1}{y^{2} - 1}

\frac{1}{\frac{11 \pm 2 i}{5} - 1} = \frac{1}{\frac{6 \pm 2 i}{5}} = \frac{5}{6 \pm 2 i} = \frac{5 (6 - 2 i)}{40}

\frac{x}{x^{2} + 1} = \frac{3 \pm i}{4}

Commented by Rasheed.Sindhi last updated on 15/Dec/22

\boldsymbol P l e a s e c h e c k m y a n s w e r f o r e r r o r s .

Commented by mnjuly1970 last updated on 15/Dec/22

x∈R ⇒ put x =tan(α) and solve answer is ((1−(√7) )/4)

x \in R \Rightarrow p u t x = t a n (α) a n d s o l v e

a n s w e r i s \frac{1 - \sqrt{7}}{4}

Commented by mr W last updated on 15/Dec/22

then something is wrong in question. for x∈R, i think (1/(1+(√(1−x+x^2 ))))+(x/(1+(√(1+x+x^2 ))))=1 has no solution.

t h e n s o m e t h i n g i s w r o n g i n q u e s t i o n .

f o r x \in R, i t h i n k

\frac{1}{1 + \sqrt{1 - x + x^{2}}} + \frac{x}{1 + \sqrt{1 + x + x^{2}}} = 1 h a s

n o s o l u t i o n .

Commented by ellenpaulisrae last updated on 15/Dec/22

h e l l o

Commented by Frix last updated on 15/Dec/22

Yes, there′s no real solution for the given equation.

Yes, {there}^{'} s no real solution for the given

equation .

Commented by Rasheed.Sindhi last updated on 16/Dec/22

ThanX manxsol sir! I′m going to correct my answer. The approach is correct although lengthy! I′ll wait for your simpler approach.But mean while mnjuly sir has changed his question and very sorry to say that he did this without any intimation! Anyway thanks again!

T h a n X m a n x s o l \boldsymbol s i r!

I^{'} m g o i n g t o c o r r e c t m y a n s w e r .

T h e a p p r o a c h i s c o r r e c t a l t h o u g h

l e n g t h y! I^{'} l l w a i t f o r y o u r s i m p l e r

a p p r o a c h . B u t m e a n w h i l e m n j u l y s i r

h a s c h a n g e d h i s q u e s t i o n a n d v e r y

s o r r y t o s a y t h a t h e d i d t h i s w i t h o u t

a n y \boldsymbol i n t i m a t i o n! A n y w a y than k s

again!

Commented by mnjuly1970 last updated on 16/Dec/22

you are right sir W i modifid the question..

y o u a r e r i g h t s i r W

i m o d i f i d t h e q u e s t i o n . .

Answered by Rasheed.Sindhi last updated on 16/Dec/22

Changed version of the question: (1/( 1+(√(1−x+x^( 2) )))) −(x/(1+(√(1+x+x^( 2) ))))=1 (1/(1+(√x) ((√(x+(1/x)−1)) )))−1=(x/(1+(√x) ((√(x+(1/x)+1)) ))) Let x+(1/x)+1=y^2 ⇒x+(1/x)−1=y^2 −2 ((1−1−(√x) ((√(y^2 −2)) ))/(1+(√x) ((√(y^2 −2)) )))=(x/(1+y(√x) )) ((−(√x) ((√(y^2 −2)) ))/(1+(√x) ((√(y^2 −2)) )))=(x/(1+y(√x) )) ((−1−(√x) ((√(y^2 −2)) ))/( (√x) ((√(y^2 −2)) )))=((1+y(√x) )/x) ((−1)/( (√x) ((√(y^2 −2)) ))−1=(1/x)+(y/( (√x))) (y/( (√x)))+(1/( (√x) ((√(y^2 −2)) ) ))=−1−(1/x)=−((x+1)/x) (y/( (√x)))+(1/( (√x) ((√(y^2 −2)) ) ))=−(((x+1)/x)) ((y(√(y^2 −2)) +1)/( (√(y^2 −2))))=−((x+1)/( (√x))) (((y(√(y^2 −2)) +1)/( (√(y^2 −2)))))^2 =(−((x+1)/( (√x))))^2 ((y^2 (y^2 −2)+1+2y(√(y^2 −2)))/(y^2 −2))=((x^2 +2x+1)/x) ((y^4 −2y^2 +1+2y(√(y^2 −2)))/(y^2 −2))=x+(1/x)+2=x+(1/x)+1+1 y^4 −2y^2 +1+2y(√(y^2 −2)) =(y^2 +1)(y^2 −2) y^4 −2y^2 +1+2y(√(y^2 −2)) =y^4 −y^2 −2 2y(√(y^2 −2)) =y^2 −3 4y^2 (y^2 −2)=y^4 −6y^2 +9 4y^4 −8y^2 −y^4 +6y^2 −9=0 3y^4 −2y^2 −9=0 y^2 =((2±(√(4+108)))/6)=((2±(√(112)))/6)=((1±2(√7))/3) ∵ y^2 is non-negative ∴ y^2 =((1+2(√7))/3) (x/(x^( 2) +1))=(1/(x+(1/x)+1−1))=(1/(y^2 −1)) =(1/(((1+2(√7))/3)−1))=(3/(−2+2(√7)))∙((−2−2(√7))/(−2−2(√7))) ((3(−2+2(√7) ))/(4−28))y=((6(−1+(√7) ))/(−24))=((1−(√7))/4)

Changed version of the question :

\frac{1}{1 + \sqrt{1 - x + x^{2}}} - \frac{x}{1 + \sqrt{1 + x + x^{2}}} = 1

\frac{1}{1 + \sqrt{x} (\sqrt{x + \frac{1}{x} - 1})} - 1 = \frac{x}{1 + \sqrt{x} (\sqrt{x + \frac{1}{x} + 1})}

L e t x + \frac{1}{x} + 1 = y^{2} \Rightarrow x + \frac{1}{x} - 1 = y^{2} - 2

\frac{1 - 1 - \sqrt{x} (\sqrt{y^{2} - 2})}{1 + \sqrt{x} (\sqrt{y^{2} - 2})} = \frac{x}{1 + y \sqrt{x}}

\frac{- \sqrt{x} (\sqrt{y^{2} - 2})}{1 + \sqrt{x} (\sqrt{y^{2} - 2})} = \frac{x}{1 + y \sqrt{x}}

\frac{- 1 - \sqrt{x} (\sqrt{y^{2} - 2})}{\sqrt{x} (\sqrt{y^{2} - 2})} = \frac{1 + y \sqrt{x}}{x}

\frac{- 1}{\sqrt{x} (\sqrt{y^{2} - 2}} - 1 = \frac{1}{x} + \frac{y}{\sqrt{x}}

\frac{y}{\sqrt{x}} + \frac{1}{\sqrt{x} (\sqrt{y^{2} - 2})} = - 1 - \frac{1}{x} = - \frac{x + 1}{x}

\frac{y}{\sqrt{x}} + \frac{1}{\sqrt{x} (\sqrt{y^{2} - 2})} = - (\frac{x + 1}{x})

\frac{y \sqrt{y^{2} - 2} + 1}{\sqrt{y^{2} - 2}} = - \frac{x + 1}{\sqrt{x}}

{(\frac{y \sqrt{y^{2} - 2} + 1}{\sqrt{y^{2} - 2}})}^{2} = {(- \frac{x + 1}{\sqrt{x}})}^{2}

\frac{y^{2} (y^{2} - 2) + 1 + 2 y \sqrt{y^{2} - 2}}{y^{2} - 2} = \frac{x^{2} + 2 x + 1}{x}

\frac{y^{4} - 2 y^{2} + 1 + 2 y \sqrt{y^{2} - 2}}{y^{2} - 2} = x + \frac{1}{x} + 2 = x + \frac{1}{x} + 1 + 1

y^{4} - 2 y^{2} + 1 + 2 y \sqrt{y^{2} - 2} = (y^{2} + 1) (y^{2} - 2)

y^{4} - 2 y^{2} + 1 + 2 y \sqrt{y^{2} - 2} = y^{4} - y^{2} - 2

2 y \sqrt{y^{2} - 2} = y^{2} - 3

4 y^{2} (y^{2} - 2) = y^{4} - 6 y^{2} + 9

4 y^{4} - 8 y^{2} - y^{4} + 6 y^{2} - 9 = 0

3 y^{4} - 2 y^{2} - 9 = 0

y^{2} = \frac{2 \pm \sqrt{4 + 108}}{6} = \frac{2 \pm \sqrt{112}}{6} = \frac{1 \pm 2 \sqrt{7}}{3}

∵ y^{2} i s n o n - n e g a t i v e

∴ y^{2} = \frac{1 + 2 \sqrt{7}}{3}

\frac{x}{x^{2} + 1} = \frac{1}{x + \frac{1}{x} + 1 - 1} = \frac{1}{y^{2} - 1}

= \frac{1}{\frac{1 + 2 \sqrt{7}}{3} - 1} = \frac{3}{- 2 + 2 \sqrt{7}} \cdot \frac{- 2 - 2 \sqrt{7}}{- 2 - 2 \sqrt{7}}

\frac{3 (- 2 + 2 \sqrt{7})}{4 - 28} y = \frac{6 (- 1 + \sqrt{7})}{- 24} = \frac{1 - \sqrt{7}}{4}

Commented by mnjuly1970 last updated on 16/Dec/22

very nice solution sir Rasheed .thanks alot for your effort. that was my mistake⟨typo⟩ i modified it.

v e r y n i c e s o l u t i o n

s i r R a s h e e d . t h a n k s a l o t

f o r y o u r e f f o r t .

t h a t w a s m y m i s t a k e ⟨ t y p o ⟩

i m o d i f i e d i t .

Answered by Rasheed.Sindhi last updated on 17/Dec/22

Simpler way (1/( 1+(√(1−x+x^( 2) )))) −(x/(1+(√(1+x+x^( 2) ))))=1 (x/(x^( 2) +1))=? (x/(x^( 2) +1))=(1/y)⇒x^2 +1=xy (1/( 1+(√(xy−x)))) −(x/(1+(√(xy+x))))=1 (1/( 1+(√x)(√(y−1)))) −(x/(1+(√x)(√(y+1))))=1 (1/( 1+(√x)(√(y−1)))) −1=(x/(1+(√x)(√(y+1)))) ((−(√x)(√(y−1)))/( 1+(√x)(√(y−1))))=(x/(1+(√x)(√(y+1)))) ((−1−(√x)(√(y−1)))/( (√x)(√(y−1))))=((1+(√x)(√(y+1)))/x) ((−1)/( (√x)(√(y−1))))−1=(1/x)+((√(y+1))/( (√x))) ((√(y+1))/( (√x)))+(1/( (√x)(√(y−1))))=−1−(1/x) (((√(y+1)) (√(y−1)) +1)/( (√x)(√(y−1))))=((−x−1)/x) ((((√(y+1)) (√(y−1)) +1)/( (√x)(√(y−1)))))^2 =(((−x−1)/x))^2 ((y^2 −1+1+2(√(y^2 −1)) )/(x(y−1)))=((x^2 +2x+1)/x^2 ) ((y^2 +2(√(y^2 −1)) )/(y−1))=((x^2 +2x+1)/x)=((xy+2x)/x)=y+2 y^2 +2(√(y^2 −1)) =(y+2)(y−1)=y^2 +y−2 2(√(y^2 −1)) =y−2 4y^2 −4=y^2 −4y+4 3y^2 +4y−8=0 y=((−4±(√(16+96)))/6)=((−4±(√(112)))/6) y=((−4±4(√7))/6)=((−2±2(√7))/3) ∵((−2+2(√7))/3) is extraneous root ∴ y=((−2−2(√7))/3) (x/(x^( 2) +1))=(1/y)=(3/(−2−2(√7)))∙((−2+2(√7))/(−2+2(√7))) =((3(−2+2(√7) ))/(4−28))=((6(−1+(√7) ))/(−24)) =((1−(√7))/4)

\boldsymbol S i m p l e r \boldsymbol w a y

\frac{1}{1 + \sqrt{1 - x + x^{2}}} - \frac{x}{1 + \sqrt{1 + x + x^{2}}} = 1

\frac{x}{x^{2} + 1} = ?

\frac{x}{x^{2} + 1} = \frac{1}{y} \Rightarrow x^{2} + 1 = x y

\frac{1}{1 + \sqrt{x y - x}} - \frac{x}{1 + \sqrt{x y + x}} = 1

\frac{1}{1 + \sqrt{x} \sqrt{y - 1}} - \frac{x}{1 + \sqrt{x} \sqrt{y + 1}} = 1

\frac{1}{1 + \sqrt{x} \sqrt{y - 1}} - 1 = \frac{x}{1 + \sqrt{x} \sqrt{y + 1}}

\frac{- \sqrt{x} \sqrt{y - 1}}{1 + \sqrt{x} \sqrt{y - 1}} = \frac{x}{1 + \sqrt{x} \sqrt{y + 1}}

\frac{- 1 - \sqrt{x} \sqrt{y - 1}}{\sqrt{x} \sqrt{y - 1}} = \frac{1 + \sqrt{x} \sqrt{y + 1}}{x}

\frac{- 1}{\sqrt{x} \sqrt{y - 1}} - 1 = \frac{1}{x} + \frac{\sqrt{y + 1}}{\sqrt{x}}

\frac{\sqrt{y + 1}}{\sqrt{x}} + \frac{1}{\sqrt{x} \sqrt{y - 1}} = - 1 - \frac{1}{x}

\frac{\sqrt{y + 1} \sqrt{y - 1} + 1}{\sqrt{x} \sqrt{y - 1}} = \frac{- x - 1}{x}

{(\frac{\sqrt{y + 1} \sqrt{y - 1} + 1}{\sqrt{x} \sqrt{y - 1}})}^{2} = {(\frac{- x - 1}{x})}^{2}

\frac{y^{2} - 1 + 1 + 2 \sqrt{y^{2} - 1}}{x (y - 1)} = \frac{x^{2} + 2 x + 1}{x^{2}}

\frac{y^{2} + 2 \sqrt{y^{2} - 1}}{y - 1} = \frac{x^{2} + 2 x + 1}{x} = \frac{x y + 2 x}{x} = y + 2

y^{2} + 2 \sqrt{y^{2} - 1} = (y + 2) (y - 1) = y^{2} + y - 2

2 \sqrt{y^{2} - 1} = y - 2

4 y^{2} - 4 = y^{2} - 4 y + 4

3 y^{2} + 4 y - 8 = 0

y = \frac{- 4 \pm \sqrt{16 + 96}}{6} = \frac{- 4 \pm \sqrt{112}}{6}

y = \frac{- 4 \pm 4 \sqrt{7}}{6} = \frac{- 2 \pm 2 \sqrt{7}}{3}

∵ \frac{- 2 + 2 \sqrt{7}}{3} i s e x t r a n e o u s r o o t

∴ y = \frac{- 2 - 2 \sqrt{7}}{3}

\frac{x}{x^{2} + 1} = \frac{1}{y} = \frac{3}{- 2 - 2 \sqrt{7}} \cdot \frac{- 2 + 2 \sqrt{7}}{- 2 + 2 \sqrt{7}}

= \frac{3 (- 2 + 2 \sqrt{7})}{4 - 28} = \frac{6 (- 1 + \sqrt{7})}{- 24}

= \frac{1 - \sqrt{7}}{4}

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