Today we will learn how to quickly square large expressions without a calculator. By large, I mean numbers ranging from ten to one hundred. Large expressions are extremely rare in real problems, and you already know how to count values less than ten, because this is a regular multiplication table. The material in today's lesson will be useful to fairly experienced students, because beginner students simply will not appreciate the speed and effectiveness of this technique.
First, let's figure out what we're talking about in general. As an example, I propose to construct an arbitrary numerical expression, as we usually do. Let's say 34. We raise it by multiplying it by itself with a column:
\[((34)^(2))=\times \frac(34)(\frac(34)(+\frac(136)(\frac(102)(1156))))\]
1156 is the square 34.
The problem with this method can be described in two points:
1) it requires written documentation;
2) it is very easy to make a mistake during the calculation process.
Today we will learn how to quickly multiply without a calculator, orally and with virtually no mistakes.
So let's get started. To work, we need the formula for the square of the sum and difference. Let's write them down:
\[(((a+b))^(2))=((a)^(2))+2ab+((b)^(2))\]
\[(((a-b))^(2))=((a)^(2))-2ab+((b)^(2))\]
What does this give us? The fact is that any value in the range from 10 to 100 can be represented as the number $a$, which is divisible by 10, and the number $b$, which is the remainder of division by 10.
For example, 28 can be represented as follows:
\[\begin(align)& ((28)^(2)) \\& 20+8 \\& 30-2 \\\end(align)\]
We present the remaining examples in the same way:
\[\begin(align)& ((51)^(2)) \\& 50+1 \\& 60-9 \\\end(align)\]
\[\begin(align)& ((42)^(2)) \\& 40+2 \\& 50-8 \\\end(align)\]
\[\begin(align)& ((77)^(2)) \\& 70+7 \\& 80-3 \\\end(align)\]
\[\begin(align)& ((21)^(2)) \\& 20+1 \\& 30-9 \\\end(align)\]
\[\begin(align)& ((26)^(2)) \\& 20+6 \\& 30-4 \\\end(align)\]
\[\begin(align)& ((39)^(2)) \\& 30+9 \\& 40-1 \\\end(align)\]
\[\begin(align)& ((81)^(2)) \\& 80+1 \\& 90-9 \\\end(align)\]
What does this idea tell us? The fact is that with a sum or a difference, we can apply the calculations described above. Of course, to shorten the calculations, for each element you should choose the expression with the smallest second term. For example, from the options $20+8$ and $30-2$, you should choose the option $30-2$.
We similarly select options for the remaining examples:
\[\begin(align)& ((28)^(2)) \\& 30-2 \\\end(align)\]
\[\begin(align)& ((51)^(2)) \\& 50+1 \\\end(align)\]
\[\begin(align)& ((42)^(2)) \\& 40+2 \\\end(align)\]
\[\begin(align)& ((77)^(2)) \\& 80-3 \\\end(align)\]
\[\begin(align)& ((21)^(2)) \\& 20+1 \\\end(align)\]
\[\begin(align)& ((26)^(2)) \\& 30-4 \\\end(align)\]
\[\begin(align)& ((39)^(2)) \\& 40-1 \\\end(align)\]
\[\begin(align)& ((81)^(2)) \\& 80+1 \\\end(align)\]
Why should we strive to reduce the second term when multiplying quickly? It's all about the initial calculations of the square of the sum and the difference. The fact is that the term $2ab$ with a plus or a minus is the most difficult to calculate when solving real problems. And if the factor $a$, a multiple of 10, is always multiplied easily, then with the factor $b$, which is a number ranging from one to ten, many students regularly have difficulties.
\[{{28}^{2}}={{(30-2)}^{2}}=200-120+4=784\]
\[{{51}^{2}}={{(50+1)}^{2}}=2500+100+1=2601\]
\[{{42}^{2}}={{(40+2)}^{2}}=1600+160+4=1764\]
\[{{77}^{2}}={{(80-3)}^{2}}=6400-480+9=5929\]
\[{{21}^{2}}={{(20+1)}^{2}}=400+40+1=441\]
\[{{26}^{2}}={{(30-4)}^{2}}=900-240+16=676\]
\[{{39}^{2}}={{(40-1)}^{2}}=1600-80+1=1521\]
\[{{81}^{2}}={{(80+1)}^{2}}=6400+160+1=6561\]
So in three minutes we did the multiplication of eight examples. That's less than 25 seconds per expression. In reality, after a little practice, you will count even faster. It will take you no more than five to six seconds to calculate any two-digit expression.
But that's not all. For those to whom the technique shown does not seem fast enough and cool enough, I propose an even faster method of multiplication, which, however, does not work for all tasks, but only for those that differ by one from multiples of 10. In our lesson there are four such values: 51, 21, 81 and 39.
It would seem much faster; we already count them in literally a couple of lines. But, in fact, it is possible to speed up, and this is done as follows. We write down the value that is a multiple of ten, which is closest to what we need. For example, let's take 51. Therefore, to begin with, let's build fifty:
\[{{50}^{2}}=2500\]
Multiples of ten are much easier to square. And now we simply add fifty and 51 to the original expression. The answer will be the same:
\[{{51}^{2}}=2500+50+51=2601\]
And so with all numbers that differ by one.
If the value we are looking for is greater than the one we are counting, then we add numbers to the resulting square. If the desired number is smaller, as in the case of 39, then when performing the action, you need to subtract the value from the square. Let's practice without using a calculator:
\[{{21}^{2}}=400+20+21=441\]
\[{{39}^{2}}=1600-40-39=1521\]
\[{{81}^{2}}=6400+80+81=6561\]
As you can see, in all cases the answers are the same. Moreover, this technique is applicable to any adjacent values. For example:
\[\begin(align)& ((26)^(2))=625+25+26=676 \\& 26=25+1 \\\end(align)\]
In this case, we do not need to remember the calculations of the squares of the sum and difference and use a calculator. The speed of work is beyond praise. Therefore, remember, practice and use in practice.
Key points
Using this technique, you can easily multiply any natural numbers ranging from 10 to 100. Moreover, all calculations are performed orally, without a calculator and even without paper!
First, remember the squares of values that are multiples of 10:
\[\begin(align)& ((10)^(2))=100,((20)^(2))=400,((30)^(2))=900,..., \\ & ((80)^(2))=6400,((90)^(2))=8100. \\\end(align)\]
\[\begin(align)& ((34)^(2))=(((30+4))^(2))=((30)^(2))+2\cdot 30\cdot 4+ ((4)^(2))= \\& =900+240+16=1156; \\\end(align)\]
\[\begin(align)& ((27)^(2))=(((30-3))^(2))=((30)^(2))-2\cdot 30\cdot 3+ ((3)^(2))= \\& =900-180+9=729. \\\end(align)\]
How to count even faster
But that is not all! Using these expressions, you can instantly square numbers “adjacent” to the reference ones. For example, we know 152 (the reference value), but we need to find 142 (an adjacent number that is one less than the reference value). Let's write it down:
\[\begin(align)& ((14)^(2))=((15)^(2))-14-15= \\& =225-29=196. \\\end(align)\]
Please note: no mysticism! Squares of numbers that differ by 1 are actually obtained by multiplying the reference numbers by themselves by subtracting or adding two values:
\[\begin(align)& ((31)^(2))=((30)^(2))+30+31= \\& =900+61=961. \\\end(align)\]
Why is this happening? Let's write down the formula for the square of the sum (and difference). Let $n$ be our reference value. Then they are calculated like this:
\[\begin(align)& (((n-1))^(2))=(n-1)(n-1)= \\& =(n-1)\cdot n-(n-1 )= \\& ==((n)^(2))-n-(n-1) \\\end(align)\]
- this is the formula.
\[\begin(align)& (((n+1))^(2))=(n+1)(n+1)= \\& =(n+1)\cdot n+(n+1) = \\& =((n)^(2))+n+(n+1) \\\end(align)\]
- a similar formula for numbers greater than 1.
I hope this technique will save you time on all your high-stakes math tests and exams. And that's all for me. See you!
The square of a number is the result of a mathematical operation that raises this number to the second power, that is, multiplies this number by itself once. It is customary to designate such an operation as follows: Z2, where Z is our number, 2 is the degree of “square”. Our article will tell you how to calculate the square of a number.
Calculate the square
If the number is simple and small, then it is easy to do this either in your head, or using the multiplication table, which we all know well. For example:
42 = 4x4 = 16; 72 = 7x7 = 49; 92 = 9x9 = 81.
If the number is large or “huge”, then you can use either the table of squares that everyone learned at school, or a calculator. For example:
122 = 12x12 = 144; 172 = 17x17 = 289; 1392 = 139x139 = 19321.
Also, to obtain the required result for the two examples above, you can multiply these numbers into a column.
In order to obtain the square of any fraction, you must:
- Convert a fraction (if the fraction has an integer part or is a decimal) into an improper fraction. If the fraction is correct, then there is no need to convert anything.
- Multiply the denominator by the denominator and the numerator by the numerator of the fraction.
For example:
(3/2)2 = (3/2)x(3/2) = (3x3)/(2x2) = 9/4; (5/7)2 = (5/7)x(5/7) = (5x5)/(7x7) = 25/49; (14/17)2 = (14x14)/(17x17) = 196/289.
In any of these options, the easiest way is to use a calculator. To do this you need:
- Type a number on the keyboard
- Click on the button with the “multiply” sign
- Press the button with the equal sign
You can also always use Internet search engines, such as Google. To do this, you just need to enter the corresponding query in the search engine field and get a ready-made result.
For example: to calculate the square of the number 9.17, you need to type 9.17*9.17, or 9.17^2, or “9.17 squared” into the search engine. In any of these options, the search engine will give you the correct result - 84.0889.
Now you know how to calculate the square of any number you are interested in, be it a whole number or a fraction, whether it is large or small!
Abbreviated multiplication formulas.
Studying abbreviated multiplication formulas: the square of the sum and the square of the difference of two expressions; difference of squares of two expressions; cube of the sum and cube of the difference of two expressions; sums and differences of cubes of two expressions.
Application of abbreviated multiplication formulas when solving examples.
To simplify expressions, factor polynomials, and reduce polynomials to standard form, abbreviated multiplication formulas are used. Abbreviated multiplication formulas you need to know by heart.
Let a, b R. Then:
1. The square of the sum of two expressions is equal to the square of the first expression plus twice the product of the first expression and the second plus the square of the second expression.
(a + b) 2 = a 2 + 2ab + b 2
2. The square of the difference of two expressions is equal to the square of the first expression minus twice the product of the first expression and the second plus the square of the second expression.
(a - b) 2 = a 2 - 2ab + b 2
3. Difference of squares two expressions is equal to the product of the difference of these expressions and their sum.
a 2 - b 2 = (a -b) (a+b)
4. Cube of sum two expressions is equal to the cube of the first expression plus triple the product of the square of the first expression and the second plus triple the product of the first expression and the square of the second plus the cube of the second expression.
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
5. Difference cube two expressions is equal to the cube of the first expression minus three times the product of the square of the first expression and the second plus three times the product of the first expression and the square of the second minus the cube of the second expression.
(a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3
6. Sum of cubes two expressions is equal to the product of the sum of the first and second expressions and the incomplete square of the difference of these expressions.
a 3 + b 3 = (a + b) (a 2 - ab + b 2)
7. Difference of cubes two expressions is equal to the product of the difference of the first and second expressions by the incomplete square of the sum of these expressions.
a 3 - b 3 = (a - b) (a 2 + ab + b 2)
Application of abbreviated multiplication formulas when solving examples.
Example 1.
Calculate
a) Using the formula for the square of the sum of two expressions, we have
(40+1) 2 = 40 2 + 2 40 1 + 1 2 = 1600 + 80 + 1 = 1681
b) Using the formula for the square of the difference of two expressions, we obtain
98 2 = (100 – 2) 2 = 100 2 - 2 100 2 + 2 2 = 10000 – 400 + 4 = 9604
Example 2.
Calculate
Using the formula for the difference of the squares of two expressions, we get
Example 3.
Simplify an expression
(x - y) 2 + (x + y) 2
Let's use the formulas for the square of the sum and the square of the difference of two expressions
(x - y) 2 + (x + y) 2 = x 2 - 2xy + y 2 + x 2 + 2xy + y 2 = 2x 2 + 2y 2
Abbreviated multiplication formulas in one table:
(a + b) 2 = a 2 + 2ab + b 2
(a - b) 2 = a 2 - 2ab + b 2
a 2 - b 2 = (a - b) (a+b)
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
(a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3
a 3 + b 3 = (a + b) (a 2 - ab + b 2)
a 3 - b 3 = (a - b) (a 2 + ab + b 2)
