The roots of the equation \(x^4 - 10x^3 + 26x^2 - 10x + 1 = 0 \) are in the form \(a \pm b \sqrt{c}\) and \(d \pm \sqrt{e}\). Find a + b + c + d + e.
+26367 The roots of the equation \(x^4 - 10x^3 + 26x^2 - 10x + 1 = 0\) are in the form \(a \pm b \sqrt{c}\) and \(d \pm \sqrt{e}\).
Find \(a + b + c + d + e\).
\(\begin{array}{|rcll|} \hline \mathbf{x^4 - 10x^3 + 26x^2 - 10x + 1} &=& \mathbf{0} \quad | \quad :x^2 \\\\ x^2-10x+26-\dfrac{10}{x}+\dfrac{1}{x^2} &=& 0 \\\\ x^2+\dfrac{1}{x^2}-10x-\dfrac{10}{x}+26 &=& 0 \\\\ \mathbf{\left(x^2+\dfrac{1}{x^2}\right)-10\left(x+\dfrac{1}{x}\right) +26} &=& \mathbf{0} \\ \hline \end{array}\)
We substitute: \(\mathbf{x+\dfrac{1}{x}=y}\)
\(\begin{array}{|rcll|} \hline \left(x+\dfrac{1}{x}\right)^2 &=& y^2 \\\\ x^2+2x*\dfrac{1}{x} + \dfrac{1}{x^2} &=& y^2 \\\\ x^2+2 + \dfrac{1}{x^2} &=& y^2 \\\\ x^2+\dfrac{1}{x^2}+2 &=& y^2 \\\\ \mathbf{x^2+\dfrac{1}{x^2}} &=& \mathbf{y^2-2} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{\left(x^2+\dfrac{1}{x^2}\right)-10\left(x+\dfrac{1}{x}\right) +26} &=& \mathbf{0} \\\\ (y^2-2)-10y +26 &=& 0 \\ y^2-10y +24 &=& 0 \\ (y-6)(y-4) &=& 0 \\ \hline \end{array}\)
\(\begin{array}{|rclcrcl|} \hline && \mathbf{(y-6)(y-4)} = \mathbf{0} \\\\ x+\dfrac{1}{x} &=& 6 \quad | \quad *x && x+\dfrac{1}{x}&=& 4 \quad | \quad *x \\\\ x^2+1&=& 6x && x^2+1&=& 4x \\ x^2-6x+1&=& 0 && x^2-4x+1&=& 0 \\\\ x &=& \dfrac{6\pm\sqrt{36-4*1} }{2} && x &=& \dfrac{4\pm\sqrt{16-4*1} }{2} \\\\ x &=& \dfrac{6\pm\sqrt{32} }{2} && x &=& \dfrac{4\pm\sqrt{12} }{2} \\\\ x &=& \dfrac{6\pm\sqrt{16*2} }{2} && x &=& \dfrac{4\pm\sqrt{4*3} }{2} \\\\ x &=& \dfrac{6\pm 4\sqrt{2} }{2} && x &=& \dfrac{4\pm 2\sqrt{3} }{2} \\\\ \mathbf{x} &=& \mathbf{3\pm 2\sqrt{2}} && \mathbf{x} &=& \mathbf{2 \pm \sqrt{3}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline a \pm b \sqrt{c} &=& 3\pm 2\sqrt{2} \\ d \pm \sqrt{e} &=& 2 \pm \sqrt{3} \\ a + b + c + d + e &=& 3+2+2+2+3 \\ \mathbf{a + b + c + d + e} &=& \mathbf{12} \\ \hline \end{array}\)
