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How do you find the square of a number? What if that number has a decimal place?

What is b^2?

What is b^2?

Answer: make use of what you already know.

## Here's the equation:

b^2 = a^2 + Ξ * Ξ£

## That's a little cryptic. Here's that equation again:

destination^2 = origin^2 + difference * sum

Note: difference = (destination - origin). Going up should have a positive difference.

## Or this:

dest^2 = orig^2 + (dest - orig)(dest + orig)

## Or, for visual simplicity:

b^2 = a^2 + (b - a)(b + a)

Think: "to get from a to b".

## In other words:

To get the square of a number (b^2, like 3.5^2 = ?), you can make use of a square that you already know (a^2, like 3^2 = 9), and just add the difference times the sum of the destination and the origin. Bonus: numbers with .5's and .1's are especially easy and can be used to get numbers with .6's.

## Example: 3.5^2 = ?

Well, with our equation we can make use of 3^2:**= 3^2 + (**

__3.5__^2__-3)(__

**3.5**__+3)__

**3.5**3.5^2 = 9 + (0.5)(3 x 2 + 0.5)

3.5^2 = 9 + half of (6 + 0.5)

3.5^2 = 9 + 3 + 0.25

So: 3.5^2 =

__.__

**12.25**We got 3.5^2 by using what we already know (3^2 = 9) and then did some calculations using relatively smaller numbers and mental shortcuts.

## Ok, but why do this?

The point was to practice problem-solving. The explicit goal was to answer a practice programming question: How do you implement or "manually" compute the square*of any number? There are multiple solutions, but one solution involves:*

__root__- ) An initial educated guess (pick smaller than the number, e.g. sqrt(17) should be smaller than 17, so maybe pick 4),
- ) checking if that guess^2 is above/below the number (e.g. 4^2 is below 17),
- ) adjusting using an efficient search of numbers below/above that number (binary search for O(logN) time), and
- ) repeating steps 2 and 3 until you get an exact answer or you settle with a close-enough answer.

Our shortcut equations give us a way to mentally do step 2, especially when steps 3 and 4 give us a number with a decimal place as we get closer and closer answers.

## Proof for the curious:

Use algebra.b^2 = a^2 + (b - a)(b + a)

b^2 = a^2 + (b^2 + ab - ab - a^2), using distributive property

b^2 = a^2 + b^2 - a^2, cancelling out

b^2 = b^2

Proven.

But that doesn't really show

*how*the equation was found. Thing is, the equation is basically the answer. So how do you find the equation?

## How the equation (i.e. answer) was found:

Discovery. This is where the real problem solving is. Find patterns in the squares of numbers, and patterns in the differences between destination and origin. Hence the equations at the beginning of this post involving destination, origin, difference, and sum.In general, make use of what you already know (like simpler squares, or ones you already calculated) and build on those things to get closer to a solution. Go with what you know. Discover patterns. Problem solve. If you just need a quick answer, you can use a calculator or computer assistant. But if you want to practice solving problems, make the most of the mental toolbox you already have, and grow from there.

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Bonus: numbers with .5's and .1's are especially easy and can be used to get numbers with .6's, .4's, .9's, and even .2's.

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