Explain why each non-zero integer has two square roots but only one cube root.

We want to study the difference between square roots and cubic roots.

The origin of this is in the law of signs, we have that:

(+)×(+) = (+)

(-)×(+) = (-)

(+)×(-) = (-)

(-)×(-) = (+)

Then for square roots we can study the exponent 2.

we can see that:

2^2 = 2×2 = 4

(-2)^2 = (-2)×(-2) = 4

So for the signs, we can see that (2)^2 = (-2)^2 = 4

And this is why √4 has two solutions:

√4 = ±2

This particularly does not happen for the cube roots for the next reason:

2^3 = 2×2×2 = 8

(-2)^3 = (-2)×(-2)×(-2) = (4)×(-2) = -8

Now because we have an odd number of elements in the product, the outcome in the negative case is also negative (this happens for all odd exponents)

So in the cube-root case, we have two different roots:

∛8 = 2

∛-8 = -2

This is also interesting because yes, each cubic root has half of the number of solutions of a square root.

But we have the double of cubic roots, that is because we can use the cubic root with a negative number, while that is not the case for a square root.

If you want to learn more, you can read:

link

if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that minus times minus = plus.

so, any when we're referring to even roots like , the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.

now, that's is not true for odd roots like , because the multiplication of the negative number will not produce a valid value, let's put two examples on that.

so, when the root is an odd root, you will always get only one number that will produce the radicand.

i got 142 as my answer

step-by-step explanation: