Square of sum: Solving the problem

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll then apply the formula we mentioned and expand the parentheses in the expression in the equation:

$(x+3)^2=x^2+9 \\ x^2+2\cdot x\cdot3+3^2=x^2+9\\ x^2+6x+9=x^2+9$We'll continue and combine like terms, by moving terms around. Later - we can notice that the squared term cancels out, 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:

$x^2+6x+9=x^2+9 \\ 6x=0\hspace{8pt}\text{/}:6\\ \boxed{x=0}$Therefore, the correct answer is answer A.

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll then apply the mentioned formula and expand the parentheses in the expression in the equation:

$(x+2)^2=x^2+12 \\ x^2+2\cdot x\cdot2+2^2=x^2+12\\ x^2+4x+4=x^2+12$We'll continue and combine like terms, by moving terms around. Later - we can notice that the squared term cancels out, 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:

$x^2+4x+4=x^2+12 \\ 4x=8\hspace{8pt}\text{/}:4\\ \boxed{x=2}$Therefore, the correct answer is answer C.

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

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

$(x+1)^2=x^2+13 \\ x^2+2\cdot x\cdot1+1^2=x^2+13 \\ x^2+2x+1=x^2+13$

We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, 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:

$x^2+2x+1=x^2+13 \\ 2x=12\hspace{8pt}\text{/}:2\\ \boxed{x=6}$Therefore, the correct answer is answer B.

Let's solve the equation, first we'll simplify the algebraic expressions using the perfect square binomial formula:

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

$(x-1)^2-(x+2)^2=15 \\ x^2-2\cdot x\cdot1+1^2-(x^2+2\cdot x\cdot2+2^2)=15 \\ x^2-2x+1-(x^2+4x+4)=15\\ x^2-2x+1-x^2-4x-4=15$In the final stage, we used the distributive property to expand the parentheses,

We'll continue and combine like terms, by moving terms between sides, later - 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:

$x^2-2x+1-x^2-4x-4=15 \\ -6x=18\hspace{8pt}\text{/}:(-6)\\ \boxed{x=-3}$Therefore, the correct answer is answer B.

Let's solve the given equation:

$x^2+10x=-25$

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

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

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

We can do this using the fact that:

$25=5^2$

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 we got 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'll ask if 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.

Let's solve the equation, first we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We will then apply the mentioned formula and expand the parentheses in the expression in the equation:

$(x+3)^2=x^2+15 \\ x^2+2\cdot x\cdot3+3^2=x^2+15\\ x^2+6x+9=x^2+15$We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, 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:

$x^2+6x+9=x^2+15 \\ 6x=6\hspace{8pt}\text{/}:6\\ \boxed{x=1}$Therefore, the correct answer is answer B.

Let's solve the given equation:

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

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

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

Now we notice that the expression on the left side can be factored using the perfect square trinomial formula:

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

We can do this using the fact that:

$4=2^2$

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 we got in the last step:

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

Notice 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.

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.

Let's examine the given equation:

$(x+2)^2-12=x^2$First, let's simplify the equation, for this we'll use 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:

$(x+2)^2-12=x^2 \\ \downarrow\\ x^2+2\cdot x\cdot2+2^2-12=x^2\\ x^2+4x+4-12= x^2\\ 4x=8\hspace{6pt}\text{/}:4\\ \boxed{x=2}$Therefore, the correct answer is answer C.