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Question
$$\frac{x}{x}+2+\frac{2}{x^2}+5x+2 = \frac{3}{x}+3$$
Answer
This equation has no solution.
Explanation
$$ \begin{aligned} \frac{x}{x}+2+\frac{2}{x^2}+5x+2 &= \frac{3}{x}+3&& \text{multiply ALL terms by } \color{blue}{ xx^2 }. \\[1 em]xx^2\frac{x}{x}+xx^2\cdot2+xx^2\cdot\frac{2}{x^2}+xx^2\cdot5x+xx^2\cdot2 &= xx^2\cdot\frac{3}{x}+xx^2\cdot3&& \text{cancel out the denominators} \\[1 em]x+2x^3+\frac{2}{x^1}+5x^4+2x^3 &= 3+3x^3&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot1x+x^1\cdot2x^3+x^1\cdot\frac{2}{x^1}+x^1\cdot5x^4+x^1\cdot2x^3 &= x^1\cdot3+x^1\cdot3x^3&& \text{cancel out the denominators} \\[1 em]x^2+2x^4+2+5x^5+2x^4 &= 3x+3x^4&& \text{simplify left and right hand side} \\[1 em]5x^5+4x^4+x^2+2 &= 3x^4+3x&& \text{move all terms to the left hand side } \\[1 em]5x^5+4x^4+x^2+2-3x^4-3x &= 0&& \text{simplify left side} \\[1 em]5x^5+x^4+x^2-3x+2 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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