solving an equation

[thereddevils]

1. The problem statement, all variables and given/known data

Solve x-\sqrt{x}-6=0

2. Relevant equations



3. The attempt at a solution

This can be factorised into (\sqrt{x}-3)(\sqrt{x}+2)=0

so \sqrt{x}=3 , x=9 .

\sqrt{x}=-2 ... Why is this root not acceptable ?

I recalled that if

x^2=4 , then x=\pm 2

but if x=\sqrt{4} , then x=2

Is this true ?

[Mentallic]

Yes, it is conventional that the square root is only the positive, so while x^2=4 and therefore x=\pm 2 ... if \sqrt{x}=-2 then this cannot be solved with x=4 since \sqrt{4}=2 and not \sqrt{4}=\pm 2.

[thereddevils]

Yes, it is conventional that the square root is only the positive, so while x^2=4 and therefore x=\pm 2 ... if \sqrt{x}=-2 then this cannot be solved with x=4 since \sqrt{4}=2 and not \sqrt{4}=\pm 2.

thanks , but why is it so , since when you square x=-2 , you still get 4 ?

[HallsofIvy]

Because (-2)(-2)= (-1)(-1)(4) and (-1)(-1)= +1.

[thereddevils]

Because (-2)(-2)= (-1)(-1)(4) and (-1)(-1)= +1.

thanks i know that , i just wonder why is it conventional for the square root positive only , why isn't the negative taken into consideration in this case ?

[Gigasoft]

The square root is a function. Do you know the definition of a function?

[thereddevils]

The square root is a function. Do you know the definition of a function?

I think so , say y=\sqrt{x} , and any input would generate only one image , why cant this image be -2 instead of 2 if the input is 4

[HallsofIvy]

thanks , but why is it so , since when you square x=-2 , you still get 4 ?
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 \sqrt{x} to be the negative root but then we would have problems with "compositions" such as \sqrt{\sqrt{x}} since the square root of a negative number is not defined in the real number system.

[thereddevils]

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 \sqrt{x} to be the negative root but then we would have problems with "compositions" such as \sqrt{\sqrt{x}} since the square root of a negative number is not defined in the real number system.

thanks a lot Hallsofivy , i finally understood.