Solving an equation

1. Apr 26, 2010

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 ?

2. Apr 26, 2010

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 $\sqrt{4}=\pm 2$.

3. Apr 26, 2010

thereddevils

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

4. Apr 26, 2010

HallsofIvy

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

5. Apr 26, 2010

thereddevils

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 ?

6. Apr 26, 2010

Gigasoft

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

7. Apr 27, 2010

thereddevils

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

8. Apr 27, 2010

HallsofIvy

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.

9. Apr 27, 2010

thereddevils

thanks a lot Hallsofivy , i finally understood.

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