The formula for the difference of squares

🏆Practice square of difference

(XY)2=X22XY+Y2(X - Y)2=X2 - 2XY + Y2
This is one of the shortened multiplication formulas and it describes the square difference of two numbers.

That is, when we encounter two numbers with a minus sign between them, that is, the difference and they will be in parentheses and raised as a squared expression, we can use this formula.

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Test yourself on square of difference!

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Choose the expression that has the same value as the following:

\( (x-7)^2 \)

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Pay attention - The formula also works in non-algebraic expressions or combinations with numbers and unknowns.

Let's look at an example

(X7)2=(X-7)^2=
Here we identify two elements between which there is a minus sign and they are enclosed in parentheses and raised to the square as a single expression.
Therefore, we can use the formula for the difference of squares.
We will work according to the formula and pay attention to the minus and plus signs.
We will obtain: 
(X7)2=x214x+49(X-7)^2=x^2-14x+49
Indeed, we pronounce the same expression differently using the formula.


If you are interested in this article, you might also be interested in the following articles:

Square

The area of a square

Multiplication of the sum of two elements by the difference between them

The formula for the sum of squares

The formulas that refer to two expressions to the power of 3

In the blog of Tutorela you will find a variety of articles about mathematics.


Examples and exercises with solutions of the difference of squares formula

Exercise #1

Choose the expression that has the same value as the following:

(xy)2 (x-y)^2

Video Solution

Step-by-Step Solution

We use the abbreviated multiplication formula:

(xy)(xy)= (x-y)(x-y)=

x2xyyx+y2= x^2-xy-yx+y^2=

x22xy+y2 x^2-2xy+y^2

Answer

x22xy+y2 x^2-2xy+y^2

Exercise #2

6016y+y2=4 60-16y+y^2=-4

Video Solution

Step-by-Step Solution

Let's solve the given equation:

6016y+y2=4 60-16y+y^2=-4 First, let's arrange the equation by moving terms:

6016y+y2=46016y+y2+4=0y216y+64=0 60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0 Now, let's note that we can break down the expression on the left side using the short quadratic factoring formula:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2 This is done using the fact that:

64=82 64=8^2 So let's present the outer term on the right as a square:

y216y+64=0y216y+82=0 y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0 Now let's examine again the short factoring formula we mentioned earlier:

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 And the expression on the left side of the equation we got in the last step:

y216y+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0 Let's note that the terms y2,82 \textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2 indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, meaning the middle term in the expression (underlined):

(ab)2=a22ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 In other words - we'll ask if it's possible to present the expression on the left side of the equation as:

y216y+82=0?y22y8+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0 And indeed it holds that:

2y8=16y 2\cdot y\cdot8=16y So we can present the expression on the left side of the given equation as a difference of two squares:

y22y8+82=0(y8)2=0 \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

(y8)2=0/y8=±0y8=0y=8 (y-8)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y-8=\pm0\\ y-8=0\\ \boxed{y=8}

Let's summarize then the solution of the equation:

6016y+y2=4y216y+64=0y22y8+82=0(y8)2=0y8=0y=8 60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}

So the correct answer is answer a.

Answer

y=8 y=8

Exercise #3

(x2)2+(x3)2= (x-2)^2+(x-3)^2=

Video Solution

Step-by-Step Solution

In order to solve the question, we need to know one of the shortcut multiplication formulas:

(xy)2=x22xy+y2 (x−y)^2=x^2−2xy+y^2

We apply the formula twice:

(x2)2=x24x+4 (x-2)^2=x^2-4x+4

(x3)2=x26x+9 (x-3)^2=x^2-6x+9

Now we add the two together:

x24x+4+x26x+9= x^2-4x+4+x^2-6x+9=

2x210x+13 2 x^2-10x+13

Answer

2x210x+13 2x^2-10x+13

Exercise #4

Look at the square below:

AAABBBDDDCCCX-7

Express its area in terms of x x .

Video Solution

Step-by-Step Solution

Remember that the area of the square is equal to the side of the square raised to the 2nd power.

The formula for the area of the square is

A=L2 A=L^2

We place the data in the formula:

A=(x7)2 A=(x-7)^2

Answer

(x7)2 (x-7)^2

Exercise #5

Given the rectangle ABCD

AB=X the ratio between AB and BC is equal tox2 \sqrt{\frac{x}{2}}

We mark the length of the diagonal A A with m m

Check the correct argument:

XXXmmmAAABBBCCCDDD

Video Solution

Step-by-Step Solution

Let's find side BC

Based on what we're given:

ABBC=xBC=x2 \frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}

xBC=x2 \frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}

2x=xBC \sqrt{2}x=\sqrt{x}BC

Let's divide by square root x:

2×xx=BC \frac{\sqrt{2}\times x}{\sqrt{x}}=BC

2×x×xx=BC \frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC

Let's reduce the numerator and denominator by square root x:

2x=BC \sqrt{2}\sqrt{x}=BC

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

AB2+BC2=AC2 AB^2+BC^2=AC^2

Let's substitute what we're given:

x2+(2x)2=m2 x^2+(\sqrt{2}\sqrt{x})^2=m^2

x2+2x=m2 x^2+2x=m^2

Answer

x2+2x=m2 x^2+2x=m^2

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