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What is the result for SQRT(-4) * SQRT(-9)?

This was posted on Brilliant, tweeted, and went viral, with thousands of views. Yet, even if it looks very simple (or impossible), you need to know complex numbers to answer this question. I won't post the solution, let's see how our members  answer it, and see if the majority (among the answers) corresponds to the solution.

For other stories that went viral, click here. For mathematical challenges, click here. And here is a very difficult question: does digit 1 appear infinitely many times in the decimals of Pi? Nobody has ever answered that question with a mathematical proof. 

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Sqrt(-4) = i2

Sqrt(-9) = i3

i2 * i3 = -6. 

This doesn't appear to be tough.  Have I oversimplified it?

The other ways to go would be to keep the powers and join the bases:

sqrt(-4*-9)=sqrt(36)=6

Richard Fowler said:

Sqrt(-4) = i2

Sqrt(-9) = i3

i2 * i3 = -6. 

This doesn't appear to be tough.  Have I oversimplified it?

Miguel,

sqrt(x)*sqrt(y) = sqrt(x*y)

is true when x and y are real, but not necessarily true when they are complex.  This happens because the square root function is defined a specific way to make it single-valued (note that (-2)^2 and 2^2 are both 4, but we define sqrt(4) to be 2, not -2).  Richard Fowler's solution is correct.

Miguel Batista said:

The other ways to go would be to keep the powers and join the bases:

sqrt(-4*-9)=sqrt(36)=6

Richard Fowler said:

Sqrt(-4) = i2

Sqrt(-9) = i3

i2 * i3 = -6. 

This doesn't appear to be tough.  Have I oversimplified it?

Exponential notation gives -6:

SQRT (-2) = i2 = 2EXP ((pi/2)*i)
SQRT (-3) = i3 = 3EXP ((pi/2)*i)
(2EXP ((pi/2)*i))*(3EXP ((pi/2)*i)) =
6EXP ((pi)*i) = 6*(-1) = -6

a^(1/r) is not the inverse of a^r in the complex plane. Therefore, you cannot apply power laws to sqrt(_) as it embodies the choice of a particular branch of a^(1/2) in the complex plane, namely the one for which (-1)^(1/2) = (0,1) = i [the other choice possible is -i].

Miguel Batista said:

The other ways to go would be to keep the powers and join the bases:

sqrt(-4*-9)=sqrt(36)=6

Richard Fowler said:

Sqrt(-4) = i2

Sqrt(-9) = i3

i2 * i3 = -6. 

This doesn't appear to be tough.  Have I oversimplified it?

In fact, this is not a calculus, but an algebra problem.
It is correct to transform the question as:
X = SQRT(-4) * SQRT(-9)  or, since SQRT(a) * SQRT(b) = SQRT(a*b), after rearranging the equation:

X - SQRT(36) = 0, which we can transform by squaring both sides as:
X^2 - 36 = 0, and by factorization, since a^2 - b^2 = (a+b) * (a-b),
(X-6) * (X+6) = 0.
This clearly shows the 2 roots, i.e:
X= - 6. and X= +6 both verify the relationship.

Hope that helps...

For one, this is not a Calculus problem.  It still falls in the realm of Algebra.

I can see Sister Mary What's-her-bottom rolling her eyes now...

Also, Richard Fowler and others are correct in that simple approach.  Unless this is some kind of silly coding riddle...
Strictly speaking, handling complex numbers is done by convention.  You have to mind extra rules when including complex numbers in radical notations.

Hi Marc,

Unfortunately, the square root function is not necessarily distributive for negative reals.  Richard Fowler's answer above is the correct one.

Marc Borowczak said:

In fact, this is not a calculus, but an algebra problem.
It is correct to transform the question as:
X = SQRT(-4) * SQRT(-9)  or, since SQRT(a) * SQRT(b) = SQRT(a*b), after rearranging the equation:

X - SQRT(36) = 0, which we can transform by squaring both sides as:
X^2 - 36 = 0, and by factorization, since a^2 - b^2 = (a+b) * (a-b),
(X-6) * (X+6) = 0.
This clearly shows the 2 roots, i.e:
X= - 6. and X= +6 both verify the relationship.

Hope that helps...

What he said..

Stephen MacMinn said:

Hi Marc,

Unfortunately, the square root function is not necessarily distributive for negative reals.  Richard Fowler's answer above is the correct one.

Marc Borowczak said:

In fact, this is not a calculus, but an algebra problem.
It is correct to transform the question as:
X = SQRT(-4) * SQRT(-9)  or, since SQRT(a) * SQRT(b) = SQRT(a*b), after rearranging the equation:

X - SQRT(36) = 0, which we can transform by squaring both sides as:
X^2 - 36 = 0, and by factorization, since a^2 - b^2 = (a+b) * (a-b),
(X-6) * (X+6) = 0.
This clearly shows the 2 roots, i.e:
X= - 6. and X= +6 both verify the relationship.

Hope that helps...

Hello, 

If you consider complex numbers then both SQRT(-4)  and SQRT(-9) are not single-valued unless it is specified that we're to use only their primary roots. 

SQRT(-4) can be 2i or -2i, and SQRT(-9) can be 3i or -3i. Only the first values in each pair are primary roots.

Hence their product is not single valued as well. It can be 6 or -6.

Look at the reference at Brilliant: it is implied that the answer is only one. Therefore they consider primary roots only, and it must be 6.

No limits = not calculus / analytic. Square the expression and take the +ve and -ve branches.

According to Matlab, the answer is -6. But you probably already know that, so I'm wondering if you might be posing a more general question. I work with complex numbers (digital filter design).

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