The formula of the difference of squares - Examples, Exercises and Solutions

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

$(x-y)^2$

We use the abbreviated multiplication formula:

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

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

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

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

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

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

$(x−y)^2=x^2−2xy+y^2$

Now, we apply this property twice:

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

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

Now we add:

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

$2 x^2-10x+13$

$2x^2-10x+13$

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

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.

$y=8$

Look at the square below:

Express its area in terms of $x$.

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:

$A=(x-7)^2$

$(x-7)^2$

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

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

$(4b-3)^2$

$16b^2-24b+9$

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

$(3x-y)^2$

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

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

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

$a^2-8a+16$

Declares the given expression as a sum

$(7b-3x)^2$

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

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

$(x-7)^2$

$x^2-14x+49$

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

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

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

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

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

$(3x-2)^2$

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

Solve using the abbreviated multiplication formula

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

$x^2+144=24x$

$x=12$

$x^2=6x-9$

$x=3$