$(X - Y)2=X2 - 2XY + Y2$
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.

### Suggested Topics to Practice in Advance

1. The formula for the sum of squares

## Examples with solutions for Square of Difference

### Exercise #1

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

$(x-y)^2$

### Step-by-Step Solution

We use the abbreviated multiplication formula:

$(x-y)(x-y)=$

$x^2-xy-yx+y^2=$

$x^2-2xy+y^2$

$x^2-2xy+y^2$

### Exercise #2

$(x-2)^2+(x-3)^2=$

### Step-by-Step Solution

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

$(x−y)^2=x^2−2xy+y^2$

We apply the formula twice:

$(x-2)^2=x^2-4x+4$

$(x-3)^2=x^2-6x+9$

Now we add the two together:

$x^2-4x+4+x^2-6x+9=$

$2 x^2-10x+13$

$2x^2-10x+13$

### Exercise #3

$60-16y+y^2=-4$

### Step-by-Step Solution

Let's solve the given equation:

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

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

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

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

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

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

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0$Let's note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2$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):

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

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

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

$\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0$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:

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

Let's summarize then the solution of the equation:

$60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}$

So the correct answer is answer a.

$y=8$

### Exercise #4

Look at the square below:

Express its area in terms of $x$.

### 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

$A=L^2$

We place the data in the formula:

$A=(x-7)^2$

$(x-7)^2$

### Exercise #5

Given the rectangle ABCD

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

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

Check the correct argument:

### Step-by-Step Solution

Let's find side BC

Based on what we're given:

$\frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}$

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

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

Let's divide by square root x:

$\frac{\sqrt{2}\times x}{\sqrt{x}}=BC$

$\frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC$

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

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

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

$AB^2+BC^2=AC^2$

Let's substitute what we're given:

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

$x^2+2x=m^2$

$x^2+2x=m^2$

### Exercise #6

Given the rectangle ABCD

Triangular area DEC equal to S

Expresses the square of the difference of the sides of the rectangle

band means of X, Y and S

### Step-by-Step Solution

Since we are given the length and width, we will substitute them according to the formula:

$(AD-AB)^2=(x-y)^2$

$x^2-2xy+y^2$

The height is equal to side AD, meaning both are equal to X

Let's calculate the area of triangle DEC:

$S=\frac{xy}{2}$

$x=\frac{2S}{y}$

$y=\frac{2S}{x}$

$xy=2S$

Let's substitute the given data into the formula above:

$(\frac{2S}{y})^2-2\times2S+(\frac{25}{x})^2$

$\frac{4S^2}{y^2}-4S+\frac{4S^2}{x^2}$

$4S=(\frac{S^2}{y}+\frac{S^2}{x}-1)$

$(x-y)^2=4s\lbrack\frac{s}{y^2}+\frac{s}{x^2}-1\rbrack$

### Exercise #7

Declares the given expression as a sum

$(7b-3x)^2$

### Video Solution

$49b^2-42bx+9x^2$

### Exercise #8

$(4b-3)(4b-3)$

Write the above as a power expression and as a summation expression.

### Video Solution

$(4b-3)^2$

$16b^2-24b+9$

### Exercise #9

Rewrite the following expression as an addition and as a multiplication:

$(3x-y)^2$

### Video Solution

$9x^2-6xy+y^2$

$(3x-y)(3x-y)$

### Exercise #10

$(a-4)(a-4)=\text{?}$

### Video Solution

$a^2-8a+16$

### Exercise #11

$(x^2-6)^2=$

### Video Solution

$x^4-12x^2+36$

### Exercise #12

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

$(x-7)^2$

### Video Solution

$x^2-14x+49$

### Exercise #13

$(x-x^2)^2=$

### Video Solution

$x^4-2x^3+x^2$

### Exercise #14

$9x^2-12x+4=$

### Video Solution

$(3x-2)^2$

### Exercise #15

$x^2-2x+1=9$

Solve using the abbreviated multiplication formula

### Video Solution

$x=-2$ o $x=4$