Find the square root of (-2-3)^2?

[parshyaa]

Root of (-2-3) ^2 = -5 ( because root of squared number is the number itself) but alsoo square of (-2-3) is 25 and its root is (+5) /(-5). Therefore what is the correct answer and reason . I think it is -5(google answer is Also -5) but I don't have any reason. Please help me

 

[mfb]

( because root of squared number is the number itself)

It is not, as your example shows. It is the magnitude of the number.

 

[Delta2]

Both answers 5 and -5 are correct for the square root of (-5)^2.

But to be more accurate when you want the square root of a number , you have to state if you want the negative or the positive square root.

When we just say "square root" we mean by convention the positive square root, so it is "a bit more correct" to say that the (positive) square root of (-5)^2 is 5.

 

[micromass]

Both answers 5 and -5 are correct for the square root of (-5)^2.

No. This is very wrong. The square root of any number is positive. So the square root of ##(-5)^2## is ##5##.

 

[Svein]

No. This is very wrong. The square root of any number is positive. So the square root of ##(-5)^2## is ##5##.

As long as you are in the real domain, yes. In the complex domain both +5 and -5 are the square roots of 25 (since there are no "positive numbers").

 

[micromass]

As long as you are in the real domain, yes. In the complex domain both +5 and -5 are the square roots of 25 (since there are no "positive numbers").

This is a common definition of the square root in complex numbers, but I don't necessarily agree with it. The problem is that it would make the square root no longer a function, which is undesirable. This is usually fixed by defining a principal square root which only evaluates to ##5## and which has a branch cut (in the same way, our square root in ##\mathbb{R}## is a principal square root too). A nicer solution exists when you go to Riemann surfaces though.

 

[Svein]

A nicer solution exists when you go to Riemann surfaces though.

That was in my mind, yes.

 

[parshyaa]

Okk I got it, answer is +/- 5 but we take 5 because of conventional use

 

[FactChecker]

It is standard to use the positive square root of a positive number. In complex analysis, that is called the "principle value" of the square root. The negative value will work but it is not the principle value.

EDIT: If you are doing your own work and taking a square root, you should often consider both the positive and negative values. If both might work, indicate that with ±√. If only the positive should be considered, indicate that with √. If only the negative should be considered, indicate that with -√. In all cases, √ just indicates the positive value.

 

[Svein]

This is a common definition of the square root in complex numbers, but I don't necessarily agree with it. The problem is that it would make the square root no longer a function, which is undesirable. This is usually fixed by defining a principal square root which only evaluates to ##5## and which has a branch cut (in the same way, our square root in ##\mathbb{R}## is a principal square root too). A nicer solution exists when you go to Riemann surfaces though.

But [itex] z^{2}=25\Leftrightarrow z^{2}-25=0 \Leftrightarrow (z+5)\cdot (z-5)=0[/itex]...

 

[SammyS]

Root of (-2-3) ^2 = -5 ( because root of squared number is the number itself) but also the square of (-2-3) is 25 and its root is (+5) /(-5). Therefore what is the correct answer and reason . I think it is -5(google answer is Also -5) but I don't have any reason. Please help me

If you had asked more symbolically, "What is ##\ \sqrt{(-2-3) ^2\,}\, ?\,##" then assuming your context was real rather than complex numbers, the answer would be simply, ##\ \sqrt{(-2-3) ^2\,}=5\ .\ ## In the context of real numbers, the ##\ \sqrt{\ \ } \ ## symbol represents the "principle value" of the square root, as pointed out by FactChecker and others.

Moreover, ##\ \sqrt{x^2\,}=|x| \ ## and not ##\ x\ .\ ## This is often surprising to students. So, the square root of a squared number is not necessarily the number itself.

 

[pwsnafu]

But [itex] z^{2}=25\Leftrightarrow z^{2}-25=0 \Leftrightarrow (z+5)\cdot (z-5)=0[/itex]...

"Find the square root of 25" and "Find all numbers that square to 25" are different mathematical questions. Solving ##z^{2}=25## is the latter. The former is ##z = \sqrt{25}##.