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Suppose x^2 - 13x + 1 = 0.  Find x^4 + 1/x^4.

 Nov 22, 2019
 #1
avatar+109740 
+2

x^2 - 13x  + 1  = 0

 

The sum of the roots  =13

The product of the roots  = 1

 

But....if the product of the  roots is  1.... we can call one root x  and the other   1/x

And the sum of the roots  =  x  + 1/x  =  13

 

So

 

(x)^2 - 13(x) + 1  = 0

(1/x)^2 - 13 (1/x) + 1  = 0        add these

 

 

(x)^2 + (1/x)^2 - 13 ( x + 1/x)  + 2  = 0

 

(x)^2  + (1/x)^2 -13 (13)  +  2  = 0

 

(x)^2 + (1/x)^2  - 169 + 2  =  0

 

(x)^2  + (1/x)^2  -167  = 0

 

(x)^2 + (1/x)^2  =   167      square both sides

 

(x)^4  + 2 (x)^2 * (1/x)^2  +  (1/x)^2  =  27889      simplify

 

x^4 + 2 (x^2/x^2) +  1/x^4  = 27889

 

(x)^4 +  2  +  (1/x)^4   +  2  =  27889       subtract 2 from both sides

 

x^2 +  1/x^4   = 27887

 

 

cool cool cool

 Nov 22, 2019
 #2
avatar+24430 
+3

Suppose \(x^2 - 13x + 1 = 0\)

Find \(x^4 + \dfrac{1}{x^4} \qquad x\neq 0!\) .

 

\(\begin{array}{|rcll|} \hline \mathbf{x^2 - 13x + 1} &=& \mathbf{0} \\\\ x^2+1 &=& 13x \quad | \quad :x \\\\ \dfrac{x^2+1}{x} &=& \dfrac{13x}{x} \\\\ \dfrac{x^2+1}{x} &=& 13 \\\\ \dfrac{x^2}{x} + \dfrac{1}{x} &=& 13 \\\\ \mathbf{ x + \dfrac{1}{x} } &=& \mathbf{13} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ x + \dfrac{1}{x} } &=& \mathbf{13} \quad | \quad \text{square both sides} \\\\ \left(x + \dfrac{1}{x} \right)^2 &=& 13^2 \\\\ x^2+2 * x*\dfrac{1}{x} + \dfrac{1}{x^2} &=& 169 \\\\ x^2 + \dfrac{1}{x^2} + 2 &=& 169 \\\\ x^2 + \dfrac{1}{x^2} &=& 169 -2 \\\\ \mathbf{ x^2 + \dfrac{1}{x^2} } &=& \mathbf{167} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ x^2 + \dfrac{1}{x^2} } &=& \mathbf{167}\quad | \quad \text{square both sides} \\\\ \left(x^2 + \dfrac{1}{x^2} \right)^2 &=& 167^2 \\\\ x^4+2 * x^2*\dfrac{1}{x^2} + \dfrac{1}{x^4} &=& 27889 \\\\ x^4 + \dfrac{1}{x^4} + 2 &=& 27889 \\\\ x^4 + \dfrac{1}{x^4} &=& 27889 -2 \\\\ \mathbf{ x^4 + \dfrac{1}{x^4} } &=& \mathbf{27887} \\ \hline \end{array}\)

 

laugh

 Nov 22, 2019

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