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# help

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

Nov 22, 2019

#1
+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

Nov 22, 2019
#2
+24430
+3

Suppose $$x^2 - 13x + 1 = 0$$

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

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

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

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

Nov 22, 2019