Solution 3.2:4
From Förberedande kurs i matematik 1
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- | {{ | + | Square both sides of the equation so that the root sign disappears, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2</math>}} |
+ | |||
+ | and then solve the resulting second-order equation by completing the square, | ||
+ | |||
+ | {{Displayed math||<math>\begin{align} | ||
+ | x^{2}-3x+3 &= 0\,,\\[5pt] | ||
+ | \Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt] | ||
+ | \Bigl(x-\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{12}{4} &= 0\,,\\[5pt] | ||
+ | \Bigl(x-\frac{3}{2}\Bigr)^{2} + \frac{3}{4} &= 0\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how ''x'' is chosen) so, the original root equation does not have any solutions. |
Current revision
Square both sides of the equation so that the root sign disappears,
\displaystyle 1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2 |
and then solve the resulting second-order equation by completing the square,
\displaystyle \begin{align}
x^{2}-3x+3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{12}{4} &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} + \frac{3}{4} &= 0\,\textrm{.} \end{align} |
As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how x is chosen) so, the original root equation does not have any solutions.