Square of sum: Transition between expressions

According to the shortened multiplication formula:

Since 7 and X appear twice, we raise both terms to the power:

$(7+x)^2$

Proceed to solve the given equation:

$x^2+10x=-25$

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

$x^2+10x=-25 \\ x^2+10x+25=0 \\$Note that the expression on the left side can be factored using the perfect square trinomial formula for a binomial squared:

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

As shown below:

$25=5^2$

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

$x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+10x+\textcolor{blue}{5}^2=0$

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 that we obtained in the last step:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0$

Notice that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{5}^2$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 (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 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 will query whether we can represent the expression on the left side as:

$\textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0$

And indeed it is true that:

$2\cdot x\cdot5=10x$

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

$\textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0$

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

$(x+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x+5=\pm0\\ x+5=0\\ \boxed{x=-5}$

Let's summarize the solution of the equation:

$x^2+10x=-25 \\ x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ x+5=0\\ \downarrow\\ \boxed{x=-5}$

Therefore the correct answer is answer C.

In order to solve the exercise, we will need to know the abbreviated multiplication formula:

In this exercise, we will use the formula twice:

$(x+1)^2=x^2+2x+1$

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

Now, we add:

$x^2+2x+1+x^2+4x+4=2x^2+6x+5$

x²+2x+1+x²+4x+4=
2x²+6x+5

Note that a common factor can be extracted from part of the digits: $2(x^2+3x)+5$

Proceed to solve the given equation:

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

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

$y^2+4y+2=-2 \\ y^2+4y+2+2=0 \\ y^2+4y+4=0$

Note that the expression on the left side can be factored using the perfect square trinomial formula:

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

As shown below:

$4=2^2$

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

$y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+4y+\textcolor{blue}{2}^2=0$

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 that we obtained in the last step:

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2=0$

Note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{2}^2$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):

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

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

$\textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2+\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}}+\textcolor{blue}{2}^2 =0$

And indeed it is true that:

$2\cdot y\cdot2=4y$

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

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

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:

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

Let's summarize the solution of the equation:

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

Therefore the correct answer is answer D.