2-(z-2)^2=z^2 — step-by-step solution

Let's solve in

ℝ

2 - {(z - 2)}^{2} = z^{2}

2 - {(z - 2)}^{2} - z^{2} = 0

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

{(z - 2)}^{2}

(z - 2) \cdot (z - 2)

2 - (z - 2) \cdot (z - 2) - z^{2} = 0

ⓘ Expand by multiplying each term inside the parentheses:

- (z - 2) \cdot (z - 2)

- (z \cdot z

+ z \cdot (-2)

- 2 z

- 2 \cdot (-2))

2 - (z \cdot z

+ z \cdot (-2)

- 2 z

- 2 \cdot (-2)) - z^{2} = 0

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

- (z \cdot z

+ z \cdot (-2)

- 2 z

- 2 \cdot (-2))

- z \cdot z

- z \cdot (-2)

+ 2 z

+ 2 \cdot (-2)

2

- z \cdot z

- z \cdot (-2)

+ 2 z

+ 2 \cdot (-2)

- z^{2} = 0

ⓘ Multiplying powers with the same base adds the exponents:

z \cdot z

z^{2}

■Two minus signs cancel out:

- z \cdot (-2)

2 z

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-2)

-4

2

- z^{2}

+ 2 z

+ 2 z

- 4

- z^{2} = 0

ⓘ Start from

2

and move back

4

on the number line, going below zero:

2 - 4

-2

■Like terms are grouped by adding their coefficients:

2 z + 2 z

4 z

- z^{2} - z^{2}

-2 z^{2}

- 2 - 2 z^{2} + 4 z = 0

Since

-2 \neq 0

, divide both sides by

-2

1 + z^{2} - 2 z = 0

Given the quadratic form

a z^{2} + b z + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -2

and

c = 1

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot 1

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-2)}^{2}

(-2) \cdot (-2)

4

Δ = 4 - 4

ⓘ Start from

4

and move back

4

on the number line, down to zero:

4 - 4

0

Δ = 0

Since

Δ = 0

, the equation has one root:

The quadratic formula gives

z_{0}

- \frac{b}{2 a}

\frac{- (-2)}{2 \cdot 1}

z_{0} = \frac{2}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

z_{0} = 1

▶Solution found:

z = 1