Números quadráticos: Unidades com 5 digitos

Squaring Numbers Ending in 5: Challenge Problems

By now, you have learned how to square numbers ending in 5. As a quick refresher, let's say you have a number ending in $5,$ and say the digits before the $5$ make up the number $N.$ Then the digits of the square of this number will begin with $N(N+1)$ and end with $25.$ For example, take a look at $115^2.$

Now that we have the basics down, let's learn now to tackle some harder questions.

Squaring Decimals Ending in .5

So, how would we square a decimal ending in $.5?$ It's very similar to squaring integers ending in $5!$ The only difference is that instead of tacking on a $25$ at the end, we tack on a $.25.$ For example, say we're trying to find $7.5^2.$

Step One: Identify $N$:

Here, the number before the $.5,$ i.e., the whole part of the decimal number, is $7,$ so $N=7.$

Step Two: Calculate $N(N+1)$:

Now, we multiply $N$ by $N+1,$ i.e. we multiply $7$ by $7+1:$ $$\begin{align} 7(7+1)&=7(8)\\ &=\boxed{56} \end{align}$$

Step Three: Add $.25$ to the End:

Now, we just tack on a $\boxed{.25}$ to the end of the $\boxed{56}$ to get the answer: $$\begin{align} 7.5^2&=\boxed{56}\boxed{.25}\\ &=56.25 \end{align}$$

Taking the Square Root of Numbers Ending in 25

Sometimes, instead of being asked to square numbers ending in $5,$ you will be asked to take the square root of certain whole numbers ending in $25.$ Be aware that your answer will not always be a whole number. For example, $\sqrt{125}=11.1803398875.$

However, sometimes your solution will be a whole number, and you can reverse engineer your process of squaring numbers ending in $5$ in order to find that solution! Let's see how it works by looking at a previous example, but through the reverse lens: we will be finding $\sqrt{13225}.$ We know this is $115$ from our earlier work, but we'll work through it in steps.

Step One: Identify the Number You're Taking the Square Root Of:

That's easy! It's $13225.$

Step Two: Chop Off the $25$ from the End:

After doing this, we're left with $132.$

Step Three: Find $N$:

This step is trickier. Remember how we multiplied $N$ with $N+1$ to find the beginning digits of our solution when we were squaring numbers ending in $5?$ Now we have to do the opposite $-$ we need to figure out two consecutive numbers, $N$ and $N+1,$ that multiply to $132.$ $N$ will always be the smaller number.

We know that $11$ and $12$ are consecutive numbers that multiply to $132,$ so $N=11.$ If this was a different number, say $210,$ we might have to go through some trial and error to figure out that $N$ and $N+1$ are $14$ and $15.$

Later, you will learn how to use the quadratic formula, which is a tool you could use to help you find $N.$ Here's an example of how to use the quadratic formula (and the distributive property) to find $N$ and $N+1$ that multiply to $132$ in case you already know it: $$\begin{align} N(N+1)&=132\\ N^2+N&=132, \text{ by the Distributive Property}\\ N^2+N-132&=0 \end{align}$$

Then by the quadratic formula: $$\begin{align} N&=\dfrac{-1\pm\sqrt{(1)^2-(4)(1)(-132)}}{(2)(1)}\\ &=\dfrac{-1\pm\sqrt{1-(4)(1)(-132)}}{2}\\ &=\dfrac{-1\pm\sqrt{1-(-528)}}{2}\\ &=\dfrac{-1\pm\sqrt{1+528}}{2}\\ &=\dfrac{-1\pm\sqrt{529}}{2}\\ &=\dfrac{-1\pm 23}{2}\\ &=\dfrac{-1+ 23}{2} \text{ or } \dfrac{-1- 23}{2}\\ &=\dfrac{22}{2} \text{ or } \dfrac{-24}{2}\\ &=11 \text{ or } -12\\ \end{align}$$

Since $N$ cannot be negative, and thus cannot be $-12,$ $N$ must be $11.$

Step Four: Put $N$ in Front of $5$:

Now that we have $N,$ it's easy to find our solution! We just need to put $N$ in front of the number $5.$ In our example, $N=11,$ so: $$\begin{align} \sqrt{13225}&=\boxed{11}\boxed{5}\\ &=115 \end{align}$$

Taking the Square Root of Decimals Ending in .25

The process of taking the square root of certain decimals ending in $.25$ is pretty much the same as taking the square root of certain integers ending in $25.$ Reminder: you won't always get a nice neat decimal. For example, $\sqrt{3.25}=1.80277563773.$

However, sometimes, your solution will be a decimal ending in $.5,$ and you can reverse engineer your process of squaring numbers ending in $.5$ in order to find that solution! Let's see how it works by looking at a previous example, but through the reverse lens: we will be finding $\sqrt{56.25}.$ We know this is $7.5$ from our earlier work, but we'll work through it in steps.

Step One: Identify the Number You're Taking the Square Root Of:

Simple! It's $56.25.$

Step Two: Chop off the $.25$ from the End:

After doing this, we're left with $56.$

Step Three: Find $N$:

Like the previous example with $132,$ we need to figure out two consecutive numbers, $N$ and $N+1,$ that multiply to $56.$ Remember, $N$ will always be the smaller number.

We know that $7$ and $8$ are consecutive numbers that multiply to $56,$ so $N=7.$ You might need to use the quadratic formula if the number is harder to factor.

Step Four: Put $N$ in Front of $.5$:

In our example, $N=7,$ so: $$\begin{align} \sqrt{56.25}&=\boxed{7}\boxed{.5}\\ &=7.5 \end{align}$$

(Video) How to find square of 2-digit numbers ending with 5, 4, 6 and 1

by faceprep

This video by faceprep shows you how to find the square of a two-digit number ending in $5.$

$\large{\text{Summary}}$

Say you have a two-digit number ending in $5,$ and say the first digit is $N.$ Then the digits of the square of this number will begin with $N(N+1)$ and end with $25,$ which is $5^2.$ Let's look at the examples he goes through in the video to clarify the trick.

$$\textbf{Example 1: }45^2$$

What is the digit before the $5?$ It's $4,$ so $N=4.$ Now, to find $N(N+1),$ we multiply $4$ by $4+1.$ $$\begin{align} N(N+1) &=4(4+1) \\ &= 4(5)\\ &=\boxed{20} \end{align}$$

Now we just tack on a $\boxed{25}$ to the end of the $\boxed{20}$ to get our answer:

$$\begin{align} 45^2 &= \boxed{20}\boxed{25}\\ &= \boxed{2025} \end{align}$$ $$\textbf{Example 2: }75^2$$

What is the digit before the $5?$ It's $7,$ so $N=7.$ Now, to find $N(N+1),$ we multiply $7$ by $7+1.$ $$\begin{align} N(N+1) &=7(7+1) \\ &= 7(8)\\ &=\boxed{56} \end{align}$$

Now we just tack on a $\boxed{25}$ to the end of the $\boxed{56}$ to get our answer:

$$\begin{align} 75^2 &= \boxed{56}\boxed{25}\\ &= \boxed{5625} \end{align}$$

Squaring Numbers That End in 5

Occasionally in math, there's a shortcut that actually works, no matter what. One of these shortcuts shows up when you need to find the square of a number that ends in $5$.

This applies for all numbers that end in $5$. Let's abbreviate these types of numbers as $N5$. The $N$ stands for anything that comes before the $5$.

For example, if our number is $\underline{43}5$, $N$ would be $\underline{43}$ (the stuff before the $5$).

Now that we're clear on what $N$ is, let's actually look at how the shortcut works.

Let's test it out. Don't use a calculator to solve the problems below. The whole point of the shortcut is to do this without a calculator.

Here, $N=4$. We know the last $2$ digits are going to be $25$. The other digits can be found by multiplying $(4)(4+1)$.

$4(4+1) = 4(5) = 20$

So, our number is $2025$.

Here, $N=7$. We know the last $2$ digits are going to be $25$. The other digits can be found by multiplying $(7)(7+1)$.

$7(7+1) = 7(8) = 56$

So, our number is $5625$.