Square of Difference - Examples, Exercises and Solutions

We use the abbreviated multiplication formula:

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

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

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

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:

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

$2 x^2-10x+13$

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.

Let's examine the given equation:

$(x-1)^2=x^2$First, let's simplify the equation, using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$,

We'll start by opening the parentheses on the left side using the perfect square formula and then move terms and combine like terms, in the final step we'll solve the simplified equation we get:

$(x-1)^2=x^2 \\ \downarrow\\ x^2-2\cdot x\cdot1+1^2=x^2\\ x^2-2x+1= x^2\\ -2x=-1\hspace{6pt}\text{/}:(-2)\\ \boxed{x=\frac{1}{2}}$Therefore, the correct answer is answer A.

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:

$A=(x-7)^2$

Let's examine the given equation:

$(x+3)^2=(x-3)^2$First, let's simplify the equation, for this we'll use the perfect square formula for a binomial squared:

$(a\pm b)^2=a^2\pm2ab+b^2$,

We'll start by opening the parentheses on both sides simultaneously using the perfect square formula mentioned, then we'll move terms and combine like terms, and in the final step we'll solve the simplified equation we get:

$(x+3)^2=(x-3)^2 \\ \downarrow\\ x^2+2\cdot x\cdot3+3^2= x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9= x^2-6x+9 \\ x^2+6x+9- x^2+6x-9 =0\\ 12x=0\hspace{6pt}\text{/}:12\\ \boxed{x=0}$Therefore, the correct answer is answer A.

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:

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

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

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:

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

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

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