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Find the number of real roots of \(1 + x + x^{2} + x^{3} = x^{4} + x^{5}\)

 Jun 18, 2020
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Find the number of real roots of  \(1 + x + x^{2} + x^{3} = x^{4} + x^{5}\)

 

\(\begin{array}{|lrcll|} \hline & 1 + x + x^2 + x^3 &=& x^4 + x^5 \\ & (1 + x)(x^2+1) &=& x^4(1+x) \\ & (1 + x)(x^2+1)-x^4(1+x) &=& 0 \\ & \mathbf{ (1 + x)(x^2+1-x^4) } &=& \mathbf{0} \\\\ 1) & x+1 &=& 0 \\ & \mathbf{x} &=& \mathbf{-1} \\\\ 2) & x^2+1-x^4 &=& 0 \\ & x^4-x^2-1 &=& 0 \\ & x^2 &=& \dfrac{1\pm \sqrt{1-4(-1)} }{2} \\ & \mathbf{x^2} &=& \mathbf{\dfrac{1\pm \sqrt{5} }{2}} \\\\ & x^2=\dfrac{1+\sqrt{5} }{2} &\text{or}& x^2=\dfrac{1-\sqrt{5} }{2} \quad < 0!\ \text{Complex solutions:} \\\\ & \mathbf{x=\sqrt{\dfrac{1+\sqrt{5}}{2}}} & \\\\ & \mathbf{x=-\sqrt{\dfrac{1+\sqrt{5} }{2}}} & \\\\ \hline \end{array}\)

 

Real solutions:
\(\mathbf{x=-1} \\ \mathbf{x=\sqrt{\dfrac{1+\sqrt{5}}{2}}} \\ \mathbf{x=-\sqrt{\dfrac{1+\sqrt{5} }{2}}}\)

 

laugh

 Jun 19, 2020

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