The square of a number is the result of a mathematical operation that raises that number to the second power, that is, it multiplies that 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.
Compute the square
If the number is simple and small, then it is easy to do it either in the mind, or using the multiplication table, which is well known to all of us. 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 desired result for the two examples above, you can multiply these numbers in a column.
In order to get the square of any fraction, you must:
- Convert a fraction (if the fraction has an integer part or if it is a decimal) to an improper fraction. If the fraction is correct, then nothing needs to be translated.
- 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 \u003d (14x14) / (17x17) \u003d 196/289.
In any of these options, the easiest way is to use a calculator. For this you need:
- Type a number on the keyboard
- Click on the button with the multiplication sign
- Press the button with the "equal" sign
You can also always use search engines on the Internet, such as, for example, Google. To do this, you just need to enter the appropriate 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 in the search engine 9.17 * 9.17, or 9.17 ^ 2, or "9.17 squared." 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 an integer or a fraction, big or small!
Abbreviated multiplication formulas.
Studying the formulas for abbreviated multiplication: the square of the sum and the square of the difference of two expressions; difference of squares of two expressions; the cube of the sum and the 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, factorize polynomials, and reduce polynomials to a 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 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 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 \u003d (a - b) (a + b)
4. sum cube of two expressions is equal to the cube of the first expression plus three times the square of the first expression times the second plus three times the product of the first expression times 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 of 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 by the incomplete square of the difference of these expressions.
a 3 + b 3 \u003d (a + b) (a 2 - ab + b 2)
7. Difference of cubes of 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 \u003d (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 squared difference of two expressions, we obtain
98 2 \u003d (100 - 2) 2 \u003d 100 2 - 2 100 2 + 2 2 \u003d 10000 - 400 + 4 \u003d 9604
Example 2
Calculate
Using the formula for the difference of the squares of two expressions, we obtain
Example 3
Simplify Expression
(x - y) 2 + (x + y) 2
We use the formulas for the square of the sum and the square of the difference of two expressions
(x - y) 2 + (x + y) 2 \u003d x 2 - 2xy + y 2 + x 2 + 2xy + y 2 \u003d 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 \u003d (a + b) (a 2 - ab + b 2)
a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2)
Today we will learn how to quickly square large expressions without a calculator. By large I mean numbers between ten and one hundred. Large expressions are extremely rare in real problems, and you already know how to count values \u200b\u200bless than ten, because this is a regular multiplication table. The material of today's lesson will be useful for fairly experienced students, because novice students simply will not appreciate the speed and effectiveness of this technique.
To begin with, let's understand in general what we are talking about. For example, I propose to do the construction of an arbitrary numeric expression, as we usually do. Let's say 34. We raise it by multiplying by itself with a column:
\[((34)^(2))=\times \frac(34)(\frac(34)(+\frac(136)(\frac(102)(1156))))\]
1156 is the square 34.
The problem of this method can be described in two points:
1) it requires written registration;
2) it is very easy to make a mistake in the process of calculation.
Today we will learn how to quickly multiply without a calculator, verbally and practically without errors.
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 between 10 and 100 can be represented as a number $a$, which is divisible by 10, and a number $b$, which is the remainder of division by 10.
For example, 28 can be represented as follows:
\[\begin(align)& ((28)^(2)) \\& 20+8 \\& 30-2 \\\end(align)\]
Similarly, we present the remaining examples:
\[\begin(align)& ((51)^(2)) \\& 50+1 \\& 60-9 \\\end(align)\]
\[\begin(align)& ((42)^(2)) \\& 40+2 \\& 50-8 \\\end(align)\]
\[\begin(align)& ((77)^(2)) \\& 70+7 \\& 80-3 \\\end(align)\]
\[\begin(align)& ((21)^(2)) \\& 20+1 \\& 30-9 \\\end(align)\]
\[\begin(align)& ((26)^(2)) \\& 20+6 \\& 30-4 \\\end(align)\]
\[\begin(align)& ((39)^(2)) \\& 30+9 \\& 40-1 \\\end(align)\]
\[\begin(align)& ((81)^(2)) \\& 80+1 \\& 90-9 \\\end(align)\]
What gives us such an idea? The fact is that with the sum or difference, we can apply the above calculations. Of course, in order to shorten the calculations, for each of the elements one should choose an expression with the smallest second term. For example, from the $20+8$ and $30-2$ options, you should choose the $30-2$ option.
Similarly, we choose options for other 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 one strive to reduce the second term in fast multiplication? It's all about the initial calculations of the square of the sum and difference. The fact is that the plus or minus term $2ab$ is the most difficult to calculate when solving real problems. And if the factor $a$, a multiple of 10, is always easily multiplied, then with the factor $b$, which is a number between one and 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. This is 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 or six seconds to calculate any two-digit expression.
But that's not all. For those to whom the shown technique seems not fast enough and not cool enough, I offer 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. There are four such values in our lesson: 51, 21, 81 and 39.
It would seem much faster, we already count them literally in a couple of lines. But, in fact, it is possible to accelerate, and this is done as follows. We write down the value, a multiple of ten, which is closest to the desired one. For example, let's take 51. Therefore, to begin with, we will raise fifty:
\[{{50}^{2}}=2500\]
Values that are 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 think, then we add numbers to the resulting square. If the desired number is less, as in the case of 39, then when performing the action, the value must be subtracted 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:
\[\begin(align)& ((26)^(2))=625+25+26=676 \\& 26=25+1 \\\end(align)\]
At the same time, we do not need to remember the calculations of the squares of the sum and difference at all 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 verbally, 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's not all! Using these expressions, you can instantly do the squaring of numbers that are “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). Let's write:
\[\begin(align)& ((14)^(2))=((15)^(2))-14-15= \\& =225-29=196. \\\end(align)\]
Please note: no mysticism! The squares of numbers that differ by 1 are indeed 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 count 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 the important tests and exams in mathematics. And that's all for me. See you!