Square roots of positive numbers

AI Thread Summary
For positive real numbers a and b, if λ² = ab, then λ can be both positive and negative, represented as λ = ±√(ab). The confusion arises from the interpretation of the square root; while the square root of a positive number is positive, squaring either a positive or negative number yields the same result. For example, if a = 1 and b = 4, then λ² = 4, leading to λ = ±2. This illustrates that both the positive and negative roots are valid solutions. Understanding this concept clarifies the presence of negatives in the square root equation.
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Homework Statement



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

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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.
 
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(-1)2 =1

so squaring -√(ab) will give (ab) regardless of a + or - before the square root.
 
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
 
A numerical example can illustrate this. Let a = 1 and b = 4 for instance. Then we have \lambda^{2} = 4. We realize however, that because negatives cancel upon multiplication, in fact (-2)^2 = 2^(2) = 4, and so both -2 and 2 are possible solutions.
 
Thank y'all for your help, that makes sense now. :)
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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