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If x^2 + 1/x^2 = 171, then what are the possible values of x - 1/x?

 Jun 22, 2020
 #1
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 Solve for x:

x^2+1/x^2==171

 

 

Bring x^2+1/x^2 together using the common denominator x^2:

x^4+1/x^2==171

 

 

Multiply both sides by x^2:

x^4+1==171 x^2

 

 

Subtract 171 x^2 from both sides:

x^4-171 x^2+1==0

 

 

Substitute y==x^2:

y^2-171 y+1==0

 

 

Subtract 1 from both sides:

y^2-171 y==-1

 

 

Add 29241/4 to both sides:

y^2-171 y+29241/4==29237/4

 

 

Write the left hand side as a square:

(y-171/2)^2==29237/4

 

 

Take the square root of both sides:

y-171/2==(13 Sqrt[173])/2 or y-171/2==-((13 Sqrt[173])/2)

 

 

Add 171/2 to both sides:

y==171/2+(13 Sqrt[173])/2 or y-171/2==-((13 Sqrt[173])/2)

 

 

Substitute back for y==x^2:

x^2==171/2+(13 Sqrt[173])/2 or y-171/2==-((13 Sqrt[173])/2)

 

 

Take the square root of both sides:

x==Sqrt[171/2+(13 Sqrt[173])/2] or x==-Sqrt[171/2+(13 Sqrt[173])/2] or y-171/2==-((13 Sqrt[173])/2)

 

 

171/2+(13 Sqrt[173])/2 == 169/4+(13 Sqrt[173])/2+173/4 == 169+26 Sqrt[173]+173/4 == 169+26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]+13)^2/4:

x==1/2 (Sqrt[173]+13) or x==-Sqrt[171/2+(13 Sqrt[173])/2] or y-171/2==-((13 Sqrt[173])/2)

 

 

171/2+(13 Sqrt[173])/2 == 169/4+(13 Sqrt[173])/2+173/4 == 169+26 Sqrt[173]+173/4 == 169+26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]+13)^2/4:

x==1/2 (13+Sqrt[173]) or x==-1/2 (Sqrt[173]+13) or y-171/2==-((13 Sqrt[173])/2)

 

 

Add 171/2 to both sides:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or y==171/2-(13 Sqrt[173])/2

 

 

Substitute back for y==x^2:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or x^2==171/2-(13 Sqrt[173])/2

 

 

Take the square root of both sides:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or x==Sqrt[171/2-(13 Sqrt[173])/2] or x==-Sqrt[171/2-(13 Sqrt[173])/2]

 

 

171/2-(13 Sqrt[173])/2 == 169/4-(13 Sqrt[173])/2+173/4 == 169-26 Sqrt[173]+173/4 == 169-26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]-13)^2/4:

x==1/2 (13+Sqrt[173]) or x==1/2 (-13-Sqrt[173]) or x==1/2 (Sqrt[173]-13) or x==-Sqrt[171/2-(13 Sqrt[173])/2]

 

 

171/2-(13 Sqrt[173])/2 == 169/4-(13 Sqrt[173])/2+173/4 == 169-26 Sqrt[173]+173/4 == 169-26 Sqrt[173]+(Sqrt[173])^2/4 == (Sqrt[173]-13)^2/4:

 

x==1/2 (13+Sqrt[173])     or     x==1/2 (-13-Sqrt[173])     or     x==1/2 (Sqrt[173]-13)     or     x==-1/2 (Sqrt[173]-13)

 

[1/2 (13+Sqrt[173])]  -  1 / [1/2 (13+Sqrt[173])] = 13   or    [1/2 (Sqrt[173]-13)]  -  1 / [1/2 (Sqrt[173]-13)] = -13

 Jun 22, 2020
 #2
avatar+25532 
+3

If
\(x^2 + \dfrac{1}{x^2} = 171\),
then what are the possible values of
\(x - \dfrac{1}{x}\)?

 

\(\begin{array}{|rcll|} \hline \left(x - \dfrac{1}{x}\right)^2 &=& x^2-2*x*\dfrac{1}{x}+\dfrac{1}{x^2} \\\\ &=& x^2+\dfrac{1}{x^2}-\dfrac{2x}{x} \\\\ &=& x^2+\dfrac{1}{x^2}-2 \quad | \quad \mathbf{x^2 + \dfrac{1}{x^2} = 171} \\\\ &=& 171-2 \\\\ &=& \mathbf{169} \\ \hline x - \dfrac{1}{x} &=& \pm \sqrt{169} \\\\ \mathbf{x - \dfrac{1}{x}} &=& \mathbf{\pm 13} \\ \hline \end{array}\)

 

 

laugh

 Jun 23, 2020

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