Square of sum - Examples, Exercises and Solutions

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.

We will first solve the exercise by opening the inner brackets:

(2[x+3])²

(2x+6)²

We will then use the shortcut multiplication formula:

(X+Y)²=X²+2XY+Y²

(2x+6)² = 2x² + 2x*6*2 + 6² = 2x+24x+36

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: $2(x^2+3x)+5$

In order to solve the exercise, remember the abbreviated multiplication formulas:

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

Let's start by using the property in both cases:

$(x+3)^2=x^2+6x+9$

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

We then reinsert them back into the formula as follows:

$2(x^2+6x+9)+3(x^2+4x+4)=$

$2x^2+12x+18+3x^2+12x+12=$

$5x^2+24x+30$

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 a square is equal to the side of the square squared

The formula for the area of a square is:

$S=a^2$

Now let's substitute the data into the formula:

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

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.

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.

Given that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

Given that AB equals X

We will substitute accordingly in the formula:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's focus on triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

Let's substitute the known values:

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

$x^2+x\times2=m^2$

We'll add 1 to both sides:

$x^2+2x+1=m^2+1$

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