The formula for the difference of squares

We use the abbreviated multiplication formula:

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

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

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

Let's solve the given equation:

$x^2+144=24x$

First, let's arrange the equation by moving terms:

$x^2+144=24x \\ x^2-24x+144=0$

Now let's notice that we can factor the expression on the left side using the perfect square trinomial formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We can do this using the fact that:

$144=12^2$

Therefore, we'll represent the rightmost term as a squared term:

$x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{12}^2=0$

Now let's examine again the perfect square trinomial formula 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 in the equation we got in the last step:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{12}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, 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 we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0$

And indeed it is true that:

$2\cdot x\cdot12=24x$

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-12)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-12=\pm0\\ x-12=0\\ \boxed{x=12}$

Let's summarize the solution of the equation:

$x^2+144=24x \\ x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0 \\ \downarrow\\ x-12=0\\ \downarrow\\ \boxed{x=12}$

Therefore the correct answer is answer C.

Let's solve the given equation:

$x^2=6x-9$

First, let's arrange the equation by moving terms:

$x^2=6x-9 \\ x^2-6x+9=0$

Now let's notice that we can factor the expression on the left side using the perfect square trinomial formula for a binomial squared:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We'll do this using the fact that:

$9=3^2$Therefore, we'll represent the rightmost term as a squared term:

$x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-6x+\textcolor{blue}{3}^2=0$

Now let's examine again the perfect square trinomial formula 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 in the equation we got in the last step:

$\textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{3}^2$indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor the expression in question (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0$

And indeed it is true that:

$2\cdot x\cdot3=6x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-3=\pm0\\ x-3=0\\ \boxed{x=3}$

Let's summarize the solution of the equation:

$x^2=6x-9 \\ x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0 \\ \downarrow\\ x-3=0\\ \downarrow\\ \boxed{x=3}$

Therefore the correct answer is answer C.

Let's solve the equation, first we'll simplify the algebraic expressions using the extended distribution law:

$(a+b)(c+d)=ac+ad+bc+bd$We'll use the shortened multiplication formula for a squared binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll therefore apply the law and formula mentioned and open the parentheses in the expressions in the equation:

$(x-4)^2=(x+2)(x-1) \\ x^2-2\cdot x\cdot4+4^2=x^2-x+2x-2 \\ x^2-8x+16=x^2+x-2$We'll continue and combine like terms, by moving terms between sides - we can notice that the squared term cancels out and therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2-8x+16=x^2+x-2\\ -9x=-18\hspace{8pt}\text{/}:(-9)\\ \boxed{x=2}$Therefore the correct answer is answer A.

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.