The sum of squares formula - Examples, Exercises and Solutions

Choose the expression that has the same value as the following:

$(x+3)^2$

We use the abbreviated multiplication formula:

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

$x^2+6x+9$

$x^2+6x+9$

$(7+x)(7+x)=\text{?}$

According to the shortened multiplication formula:

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

$(7+x)^2$

$(7+x)^2$

$4x^2+20x+25=$

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

Let's remember the formulas:

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

 $(x+y)^2=x^2+2xy+y^2$

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

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

Now let's try to think, what number multiplied by itself will equal 4 and what number multiplied by itself will equal 25?

The answers are respectively 2 and 5:

We will write:

$(2x+5)^2=$

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

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

$4x^2+20x+25$

That means our solution is correct.

$(2x+5)^2$

$(2\lbrack x+3\rbrack)^2=$

First, we will solve the exercise by opening the inner brackets:

(2[x+3])²

(2x+6)²

Now we will use the shortcut multiplication formula:

(X+Y)²=X²+2XY+Y²

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

$4x^2+24x+36$

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

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

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

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

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 will place them back in the formula:

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

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

$5x^2+24x+30$

$5x^2+24x+30$

Choose the expression that has the same value as the following:

$(x+y)^2$

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

$(7+8)^2=\text{?}$

$7^2+2\times7\times8+8^2$

$(a+b)^2=\text{?}$

$a^2+2ab+b^2$

$y=x^2+9x+24$

Which expression should be added to y so that:

$y=(x+5)^2$

$x+1$

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

$x^4+8x^2+16$

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

$x^2+2x^3+x^4$

Solve for x:

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

$x=0$

$(3x+4)^2=\text{?}$

$9x^2+24x+16$

$x^2+16x+64=81$

Find X

$x=-17$ o $x=1$