Square of sum - Examples, Exercises and Solutions

To solve this problem, we will use the formula for the square of a sum, which is $(a+b)^2 = a^2 + 2ab + b^2$.

Here, we identify $a = x$ and $b = y$. Thus, our expression $(x+y)^2$ can be expanded as follows:

$(x+y)^2 = x^2 + 2xy + y^2$

Now we will match this expanded form with the given choices.

• Choice 1: $x^2 + y^2$ - This is missing the middle term $2xy$.
• Choice 2: $y^2 + x^2 + 2xy$ - This matches the expanded expression.
• Choice 3: $x^2 + xy + y^2$ - The term $xy$ doesn't match the required $2xy$.
• Choice 4: $x^2 - 2xy + y^2$ - The middle term has an incorrect sign.

Therefore, the expression that is equivalent to $(x+y)^2$ is choice 2: $y^2 + x^2 + 2xy$.

We use the abbreviated multiplication formula:

$x^2+2\times x\times3+3^2=$

$x^2+6x+9$

According to the shortened multiplication formula:

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

$(7+x)^2$

In this task, we are asked to simplify the formula using the abbreviated multiplication formulas.

Let's take a look at the formulas:

$(x-y)^2=x^2-2xy+y^2$

 $(x+y)^2=x^2+2xy+y^2$

$(x+y)\times(x-y)=x^2-y^2$

Taking into account that in the given exercise there is only addition operation, the appropriate formula is the second one:

Now let us consider, what number when multiplied by itself will equal 4 and what number when multiplied by itself will equal 25?

The answers are respectively 2 and 5:

We insert these into the formula:

$(2x+5)^2=$

$(2x+5)(2x+5)=$

$2x\times2x+2x\times5+2x\times5+5\times5=$

$4x^2+20x+25$

That means our solution is correct.

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.

To solve the expression $(x^2 + 4)^2$, we will follow these steps:

• Step 1: Identify the expression as a binomial $(a + b)$, where $a = x^2$ and $b = 4$.
• Step 2: Apply the binomial square formula: $(a + b)^2 = a^2 + 2ab + b^2$.
• Step 3: Substitute $a$ and $b$ into the formula.
• Step 4: Calculate each term in the formula.
• Step 5: Simplify to arrive at the final expanded form.

Let's execute these steps:
Step 1: Identify $a = x^2$ and $b = 4$.
Step 2: Apply the formula $(a + b)^2 = a^2 + 2ab + b^2$.
Step 3: Substitute to get: $(x^2)^2 + 2(x^2)(4) + 4^2$.
Step 4: Calculate each term:
- $(x^2)^2 = x^4$,
- $2(x^2)(4) = 8x^2$,
- $4^2 = 16$.
Step 5: Combine the terms to get the expanded expression: $x^4 + 8x^2 + 16$.

Therefore, the solution to the expression is $x^4 + 8x^2 + 16$.

To solve this problem, follow these steps:

• Step 1: Identify the terms: Here, $a = x$ and $b = x^2$.
• Step 2: Apply the square of a binomial formula: $(a + b)^2 = a^2 + 2ab + b^2$.
• Step 3: Substitute the terms into the formula and simplify the resulting expression.

Now, let's work through each step:

Step 1: We have $a = x$ and $b = x^2$.
Step 2: The formula gives us: $(a + b)^2 = a^2 + 2ab + b^2$

Step 3: Substituting $a = x$ and $b = x^2$ into the formula:

This simplifies to:

Therefore, the expanded form of the expression $(x + x^2)^2$ is $\mathbf{x^2 + 2x^3 + x^4}$.

This matches with choice 3 from the options provided.

To solve this problem, we'll follow these steps:

• Identify the given expression $(a+b)^2$.
• Apply the formula for the square of a sum: $(x+y)^2 = x^2 + 2xy + y^2$.
• Substitute $x = a$ and $y = b$ into the formula and simplify.

Let's apply these steps to the expression $(a+b)^2$:
We start with the expression $(a+b)^2$. This means we are squaring the sum $a + b$.

According to the formula $(x+y)^2 = x^2 + 2xy + y^2$, we can substitute $x = a$ and $y = b$. Therefore, the expression becomes:

$(a+b)^2 = a^2 + 2ab + b^2$.

Therefore, the expanded form of the expression $(a+b)^2$ is $a^2 + 2ab + b^2$.

To solve this problem, we'll use the formula for the square of a sum.

• Step 1: Identify our variables as $a = 7$ and $b = 8$.
• Step 2: Apply the formula $(a + b)^2 = a^2 + 2ab + b^2$.
• Step 3: Substitute into the formula:
  $(7 + 8)^2 = 7^2 + 2 \times 7 \times 8 + 8^2$.
• Step 4: Express the final expression in the most informative form.

Therefore, the expanded expression for $(7 + 8)^2$ is $7^2 + 2 \times 7 \times 8 + 8^2$.

Regarding the choices provided, the correct one is Option 3: $7^2 + 2 \times 7 \times 8 + 8^2$.

Let's solve the equation $x^2 + 16x + 64 = 81$ step-by-step:

• Step 1: Recognize the Left-Hand Side as a Perfect Square
  Notice that the left-hand side, $x^2 + 16x + 64$, can be factored as $(x + 8)^2$ because $(x + 8)^2 = x^2 + 16x + 64$.
• Step 2: Set the Factored Expression Equal to 81
  We have $(x + 8)^2 = 81$.
• Step 3: Take the Square Root of Both Sides
  Taking the square root on both sides gives two possible equations: $x + 8 = 9$ and $x + 8 = -9$.
• Step 4: Solve Each Equation
  For $x + 8 = 9$:
  Subtract 8 from both sides to get $x = 1$.
  For $x + 8 = -9$:
  Subtract 8 from both sides to get $x = -17$.

Therefore, the solutions to the equation are $x = -17$ or $x = 1$.

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.

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 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:

$(x+1)^2=x^2 \\ x^2+2\cdot x\cdot1+1^2=x^2 \\ x^2+2x+1=x^2 \\$We'll continue and combine like terms, by moving terms between sides. 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:

$x^2+2x+1=x^2 \\ 2x=-1\hspace{8pt}\text{/}:2\\ \boxed{x=-\frac{1}{2}}$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 apply the mentioned formula and expand the parentheses in the expressions in the equation:

$(x+2)^2-12=x^2 \\ x^2+2\cdot x\cdot2+2^2-12=x^2 \\ x^2+4x+4-12=x^2 \\$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:

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