Solving x-\sqrt{x}-6=0: Understanding the Acceptable Roots of Equations

[thereddevils]

Homework Statement

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

Homework Equations

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 can't 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.