Square of sum: Transition between expressions

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

According to the shortened multiplication formula:

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

$(7+x)^2$

$(7+x)^2$

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

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

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

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

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

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

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

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

$9x^2+24x+16$

$x^2+10x=-25$

$x=-5$

$4x^2=12x-9$

$x=\frac{3}{2}$

$2^2+12+3^2=\text{?}$

$(2+3)^2$

Express the following exercise as a sum and as a power:

$(7b+3z)(7b+3z)=\text{?}$

$49b^2+42bz+9z^2$

$(7b+3z)^2$

Rewrite the following expression as a multiplication and as an addition:

$(a+3b)^2$

$(a+3b)(a+3b)$

$a^2+6ab+9b^2$

Solve for y:

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

$y=-2$

Solve for x:

$x^2+32x=-256$

$x=-16$