If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Square roots of perfect squares

Learn how to find the square root of perfect squares like 25, 36, and 81.
Let's start by taking a look at an example evaluating the square root of $25$:
$\sqrt{25}=\phantom{\rule{0.167em}{0ex}}?$
Step 1: Ask, "What number squared equals $25$?"
Step 2: Notice that $5$ squared equals $25$.
${5}^{2}=5×5=25$
$\sqrt{25}=5$
Here's a question to make sure you understood:
How can we be sure that $5$ is the right answer?

# Connection to a square

Finding the square root of $25$ is the same as finding the side length of a square with an area of $25$.
A square with an area of $25$ has a side length of $5$.

# Practice Set 1:

Problem 1A
${4}^{2}=$

### Reflection question

Which claim shows how square roots work?

# Practice Set 2:

Problem 2A
$\sqrt{1}=$

# Practice Set 3:

Problem 3A
$\sqrt{121}=$

## Want to join the conversation?

• Can an exponent be a negative number?
• 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.
• What happens if try to find the
the square root of an imperfect square
• 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
• 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
• 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
• So, a perfect square is basically the answer to an exponent? (ex. 2^2(4), 12^12(144))
• it is an integer with a square root that is also an integer
• How do you solve square roots that can't be squared easily?
• what i do is if the number is less then 81 then I go through all of the numbers 1-9 to see if they match up. if its greater than 144 then i just use a calculator
• How can we use square roots in life?
• There are many ways in life you can use square roots. Some of them include a mathmatician, engineer, architect, carpenter, etc.
• Would you do the same for a negative square root ?🤔
• 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.
• can a decimal have a square root
• 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.
• Math is hard!
• it is not if you listen carefully in class and practice
• How do I find the square root for bigger and not perfect numbers?
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.