Question 1

Can an exponent be a negative number?

Accepted Answer

You already know that an exponent represents the number of times you have to multiply a number by itself. A negative exponent is equivalent to the inverse of the same number with a positive exponent. There is nothing special about solving a problem that includes negative exponentials.

Question 2

What happens if try to find the 
the square root of an imperfect square

Accepted Answer

Most probably it will be an irrational number in which case we can only approximate its value.

However, sometimes we can write the radicand as a fraction of two perfect squares.
Example: √5.76 = √(576∕100) = √(144∕25) = √(12²∕5²) = 12∕5 = 2.4

Question 3

I just noticed something really interesting (I think):

If I can't remember a square of some numbers (7^2 and 8^2 can be a bit tricky for me for some reason) but I remember the square number of the root that comes before it (6 and 36 in the case I'm trying to find 7^2),
I can do 36+6 to make it into 6*7, and then add a 7 to make it into a 7*7.

I tried to play around and find a rule and I think I found the formula:
n^2 + 2n + 1 = ( n + 1 )^2
(where n is the root number of the square that you do know).

if you would visualize the numbers on a grid, the n^2 is the area of the square, while the 2n+1 is the number of the additional units the is added on the side.

It's easier to see it on a times table: When looking at 25, the number diagonally next to it is 36. If you count the 'units' (the other multiplies that are on the same axes that leads toward 36 IE: 3,6,9 up to 36 on both sides) they will be the same as the 2n+1.

I tried to find a formula for a square of a root that isn't immediately follows the root I know. example: 3^2 = 9, 5^2 = ?.
unfortunately I couldn't think on one consistent formula, because there is always a need to add more and (n+1) with more additional 1s the further the number is.

does the formula I found have a name? I'm pretty sure I wasn't the first to think of that lol

Accepted Answer

You can use the pascal's triangle I suppose

Edit1: here: https://en.wikipedia.org/wiki/Polynomial_expansion

Edit2: Sorry that is not true. That applies to expansions with higher powers, but not between different sqaures.
But maybe you can look at sum of squares where
∑x^2 = n(n+1)(2n+1)/6

Question 4

So, a perfect square is basically the answer to an exponent? (ex. 2^2(4), 12^12(144))

Accepted Answer

it is an integer with a square root that is also an integer

Question 5

How can we use square roots in life?

Accepted Answer

There are many ways in life you can use square roots. Some of them include a mathmatician, engineer, architect, carpenter, etc.

Question 6

can a decimal have a square root

Accepted Answer

Yes, decimals can have square roots.  For example, the (principal) square root of 0.64 is 0.8, since 0.8*0.8 = 0.64.

Question 7

How do I find the square root for bigger and not perfect numbers?

Accepted Answer

There are a couple of ways:

1. You can approximate the value. For instance, if I have $\sqrt{70}$, I know that it'll lie between $\sqrt{64}$ and $\sqrt{81}$. As 70 is closer to 64, $\sqrt{70}$ will be closer to $\sqrt{64}$, which is 8. And if you find $\sqrt{70}$, you'll see that it is around $8.366$, which is closer to 8 than 9.

2. You can approximate the square root function with a tangent line at a nearby point. Suppose you want $\sqrt{65}$. You already know $\sqrt{64}$. So, you can first graph $y=\sqrt{x}$, then draw a tangent at $x = 64$. Find the equation of the tangent, and substitute for $x = 65$. As the tangent approximates the square root function, you'll get an rough approximation of $\sqrt{65}$.

Using this, you can get the equation of the tangent line as $y = \frac{x}{16}+4$. If I substitute $x = 65$, I get $y = 8.0625$. Now, the actual value of $\sqrt{65}$ is $8.06225$. See how good our approximation was!

3. The final method is something called a Taylor series. It's an extension of the tangent line approximation idea. If you write the Taylor series expansion of $\sqrt{x}$ at $x = 64$ to four terms, you get $8 + \frac{x-64}{16} - \frac{(x-64)^2}{4096} + \frac{(x-64)^3}{524288}$. If I substitute $x = 65$, I'll get $8 + \frac{1}{16} - \frac{1}{4096} + \frac{1}{524288} = 8.062257$, which is an even better approximation than what we got in method 2. It gets better with more terms.

I know you aren't familiar with methods 2 and 3. But hey, if you stick with Math long enough, you'll one day come back to this comment and understand everything!

Question 8

Would you do the same for a negative square root ?🤔

Accepted Answer

You cannot find the negative square root of a number. Think of what square roots are. If we take 36 and find the square root, you'll see it's 6. Because 6 x 6 = 36. We multiply it by itself and it gives us 36.

We can't do this with negatives. Consider -36 and finding the square root of that. Which number multiplied by itself will get -36?

Well, remember the rules we have for multiplying negatives and positives. 
negative x negative = positive
negative x positive = positive
positive x positive = positive.

It can't be -6, because that will give us a positive number, instead of -36.
It can't be 6, because that will also give us a positive number instead of -36.

So, negatives can't have square roots.

You'll learn in more advanced classes what we do in these cases.

Course: 8th grade > Unit 1

Square roots of perfect squares

Connection to a square

Practice Set 1:

Reflection question

Practice Set 2:

Practice Set 3:

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