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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,

1−x=(2−x)21−x=4−4x+x2

and then solve the resulting second-order equation by completing the square,

x2−3x+3x−232−232+3x−232−49+412x−232+43=0=0=0=0.

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.

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