x-1=(2-x)^2 — step-by-step solution

1) Let's solve in

ℝ

x - 1 = {(2 - x)}^{2}

x - 1 - {(2 - x)}^{2} = 0

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

{(2 - x)}^{2}

(2 - x) \cdot (2 - x)

x - 1 - (2 - x) \cdot (2 - x) = 0

ⓘ Expand by multiplying each term inside the parentheses:

- (2 - x) \cdot (2 - x)

- (2 \cdot 2

+ 2 \cdot (- x)

- x \cdot 2

- x \cdot (- x))

x - 1 - (2 \cdot 2

+ 2 \cdot (- x)

- x \cdot 2

- x \cdot (- x)) = 0

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

- (2 \cdot 2

+ 2 \cdot (- x)

- x \cdot 2

- x \cdot (- x))

- 2 \cdot 2

- 2 \cdot (- x)

+ x \cdot 2

+ x \cdot (- x)

x

- 1

- 2 \cdot 2

- 2 \cdot (- x)

+ x \cdot 2

+ x \cdot (- x) = 0

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 2

4

■Two minus signs cancel out:

-2 \cdot (- x)

2 x

■Like terms are grouped by adding their coefficients:

x + x \cdot 2

3 x

■The minus sign is placed in front of the product:

x \cdot (- x)

- x \cdot x

3 x

- 1

- 4

+ 2 x

- x \cdot x = 0

ⓘ Start from

-1

and move back

4

more on the number line, away from zero:

- 1 - 4

-5

■Like terms are grouped by adding their coefficients:

3 x + 2 x

5 x

■Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

5 x - 5 - x^{2} = 0

- x^{2} + 5 x - 5 = 0

Since

-1 \neq 0

, divide both sides by

-1

x^{2} - 5 x + 5 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -5

and

c = 5

Δ = {(-5)}^{2} - 4 \cdot 1 \cdot 5

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

{(-5)}^{2}

(-5) \cdot (-5)

25

Δ = 25 - 4 \cdot 5

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 5

20

Δ = 25 - 20

ⓘ Start from

25

and move back

20

on the number line:

25 - 20

5

Δ = 5

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-5) - \sqrt{5}}{2 \cdot 1}

x_{1} = \frac{5 - \sqrt{5}}{2}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-5) + \sqrt{5}}{2 \cdot 1}

x_{2} = \frac{5 + \sqrt{5}}{2}

▶Solutions found:

x = {\frac{5 - \sqrt{5}}{2}, \frac{5 + \sqrt{5}}{2}}

2) The numeric value of:

x = {\frac{5 - \sqrt{5}}{2}, \frac{5 + \sqrt{5}}{2}}

equals:

x = {\approx \frac{5 - 2.2361}{2}, \approx \frac{5 + 2.2361}{2}}

x = {\approx \frac{2.7639}{2}, \approx \frac{7.2361}{2}}

▷

x = {\approx 1.3820, \approx 3.6180}