# Homework Help: Square roots of positive numbers

1. Mar 9, 2010

### bluskies

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

If a and b are positive real numbers, and $$\lambda^{2} = ab$$, then $$\lambda = \pm \sqrt{ab}$$.

2. Relevant equations

None.

3. The attempt at a solution

This is more of a conceptual question that has always escaped me. I do not understand how the square root of two positive numbers could possibly be negative. Since a and b are positive, how can there be any negatives in the square root of their product? Any guidance on this subject would be very much appreciated.

2. Mar 9, 2010

### rock.freak667

(-1)2 =1

so squaring -√(ab) will give (ab) regardless of a + or - before the square root.

3. Mar 9, 2010

### dacruick

I think you might be looking too hard into this. There are no negatives in the square root. the negative is outside it. say both a and b are 5. then lambda² = 25. lambda therefore is equal to positive root(25) or negative root(25). Or in other terms, lambda² = 5² or (-5)²

4. Mar 9, 2010

### Fightfish

A numerical example can illustrate this. Let a = 1 and b = 4 for instance. Then we have $$\lambda^{2} = 4$$. We realise however, that because negatives cancel upon multiplication, in fact $$(-2)^2 = 2^(2) = 4$$, and so both -2 and 2 are possible solutions.

5. Mar 9, 2010

### bluskies

Thank y'all for your help, that makes sense now. :)