+128406 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
+26367 Suppose \(x^2 - 13x + 1 = 0\).
Find \(x^4 + \dfrac{1}{x^4} \qquad x\neq 0!\) .
\(\begin{array}{|rcll|} \hline \mathbf{x^2 - 13x + 1} &=& \mathbf{0} \\\\ x^2+1 &=& 13x \quad | \quad :x \\\\ \dfrac{x^2+1}{x} &=& \dfrac{13x}{x} \\\\ \dfrac{x^2+1}{x} &=& 13 \\\\ \dfrac{x^2}{x} + \dfrac{1}{x} &=& 13 \\\\ \mathbf{ x + \dfrac{1}{x} } &=& \mathbf{13} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \mathbf{ x + \dfrac{1}{x} } &=& \mathbf{13} \quad | \quad \text{square both sides} \\\\ \left(x + \dfrac{1}{x} \right)^2 &=& 13^2 \\\\ x^2+2 * x*\dfrac{1}{x} + \dfrac{1}{x^2} &=& 169 \\\\ x^2 + \dfrac{1}{x^2} + 2 &=& 169 \\\\ x^2 + \dfrac{1}{x^2} &=& 169 -2 \\\\ \mathbf{ x^2 + \dfrac{1}{x^2} } &=& \mathbf{167} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \mathbf{ x^2 + \dfrac{1}{x^2} } &=& \mathbf{167}\quad | \quad \text{square both sides} \\\\ \left(x^2 + \dfrac{1}{x^2} \right)^2 &=& 167^2 \\\\ x^4+2 * x^2*\dfrac{1}{x^2} + \dfrac{1}{x^4} &=& 27889 \\\\ x^4 + \dfrac{1}{x^4} + 2 &=& 27889 \\\\ x^4 + \dfrac{1}{x^4} &=& 27889 -2 \\\\ \mathbf{ x^4 + \dfrac{1}{x^4} } &=& \mathbf{27887} \\ \hline \end{array}\)
