The discussion revolves around solving the equation x - √x - 6 = 0, specifically focusing on the acceptable roots of equations involving square roots and the conventions surrounding them.
Participants are engaging in a productive dialogue about the conventions of square roots as functions, with some clarifying that the square root function is defined to yield only non-negative outputs. There is an exploration of the implications of this definition on the roots of the equation.
There is a focus on the definition of functions and the implications of choosing positive versus negative roots, as well as the constraints of the real number system regarding square roots of negative numbers.
Mentallic said:Yes, it is conventional that the square root is only the positive, so while [itex]x^2=4[/itex] and therefore [itex]x=\pm 2[/itex] ... if [itex]\sqrt{x}=-2[/itex] then this cannot be solved with x=4 since [itex]\sqrt{4}=2[/itex] and not [itex]\sqrt{4}=\pm 2[/itex].
HallsofIvy said:Because (-2)(-2)= (-1)(-1)(4) and (-1)(-1)= +1.
Gigasoft said:The square root is a function. Do you know the definition of a function?
Sorry, I misunderstood your question. A "function" can give only one value for each value of x so we must choose either the positive or negative root0. We could define [itex]\sqrt{x}[/itex] to be the negative root but then we would have problems with "compositions" such as [itex]\sqrt{\sqrt{x}}[/itex] since the square root of a negative number is not defined in the real number system.thereddevils said:thanks , but why is it so , since when you square x=-2 , you still get 4 ?
HallsofIvy said:Sorry, I misunderstood your question. A "function" can give only one value for each value of x so we must choose either the positive or negative root0. We could define [itex]\sqrt{x}[/itex] to be the negative root but then we would have problems with "compositions" such as [itex]\sqrt{\sqrt{x}}[/itex] since the square root of a negative number is not defined in the real number system.