x(x+12)=(x+6)^2 — step-by-step solution

Let's solve in

ℝ

x \cdot (x + 12) = {(x + 6)}^{2}

x \cdot (x + 12) - {(x + 6)}^{2} = 0

ⓘ Expand by multiplying each term inside the parentheses:

x \cdot (x + 12)

x^{2} + x \cdot 12

■Raising a sum to an integer power is multiplying it by itself:

{(x + 6)}^{2}

(x + 6) \cdot (x + 6)

x^{2} + x \cdot 12 - (x + 6) \cdot (x + 6) = 0

ⓘ Expand by multiplying each term inside the parentheses:

- (x + 6) \cdot (x + 6)

- (x \cdot x

+ x \cdot 6

+ 6 x

+ 6 \cdot 6)

x^{2} + x \cdot 12 - (x \cdot x

+ x \cdot 6

+ 6 x

+ 6 \cdot 6) = 0

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (x \cdot x

+ x \cdot 6

+ 6 x

+ 6 \cdot 6)

- x \cdot x

- x \cdot 6

- 6 x

- 6 \cdot 6

x^{2}

+ x \cdot 12

- x \cdot x

- x \cdot 6

- 6 x

- 6 \cdot 6 = 0

ⓘ Like terms are grouped by adding their coefficients:

x \cdot 12

- x \cdot x

- x \cdot 6

- 6 x

0 - x^{2}

x^{2} - x^{2}

0

■Multiplications are done before additions and subtractions:

6 \cdot 6

36

0 + 0 - 36 = 0

ⓘ Adding 0 changes nothing:

0 + 0 - 36

-36

-36 = 0

The statement

-36 = 0

is false, whatever the value of

x

▶Solution:

x \in \emptyset