(x+1)^2=-2(x-1)^2 — step-by-step solution

Let's solve in

ℝ

{(x + 1)}^{2} = -2 \cdot {(x - 1)}^{2}

{(x + 1)}^{2} + 2 \cdot {(x - 1)}^{2} = 0

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

{(x + 1)}^{2}

(x + 1) \cdot (x + 1)

{(x - 1)}^{2}

(x - 1) \cdot (x - 1)

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

ⓘ Expand by multiplying each term inside the parentheses:

(x + 1) \cdot (x + 1)

x \cdot x

+ x \cdot 1

+ 1 \cdot x

+ 1 \cdot 1

2 \cdot (x - 1)

2 x - 2

x \cdot x

+ x \cdot 1

+ 1 \cdot x

+ 1 \cdot 1

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

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Multiplications are done before additions and subtractions:

1 \cdot 1

1

x^{2}

+ x

+ x

+ 1

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

ⓘ Like terms are grouped by adding their coefficients:

x + x

2 x

x^{2}

+ 2 x

+ 1

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

ⓘ Expand by multiplying each term inside the parentheses:

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

2 \cdot x \cdot x

+ 2 x \cdot (-1)

- 2 x

- 2 \cdot (-1)

x^{2}

+ 2 x

+ 1

+ 2 \cdot x \cdot x

+ 2 x \cdot (-1)

- 2 x

- 2 \cdot (-1) = 0

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■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 (-1)

-2

■Like terms are grouped by adding their coefficients:

2 x - 2 x

0

■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

-2 \cdot (-1)

2

x^{2}

+ 0

+ 1

+ 2 x^{2}

- 2 x

+ 2 = 0

ⓘ Adding 0 changes nothing:

0 + 1

1

■Start from

1

and move forward

2

on the number line:

1 + 2

3

■Like terms are grouped by adding their coefficients:

x^{2} + 2 x^{2}

3 x^{2}

3 x^{2} + 3 - 2 x = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 3

b = -2

and

c = 3

Δ = {(-2)}^{2} - 4 \cdot 3 \cdot 3

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

{(-2)}^{2}

(-2) \cdot (-2)

4

■Multiplications are done before additions and subtractions:

4 \cdot 3

12

Δ = 4 - 12 \cdot 3

ⓘ Multiplications are done before additions and subtractions:

12 \cdot 3

(10 + 2) \cdot 3

10 \cdot 3 + 2 \cdot 3

30 + 6

36

Δ = 4 - 36

ⓘ Start from

4

and move back

36

on the number line, going below zero:

4 - 36

-32

Δ = -32

Δ < 0

, the equation has no real root.

▶Solution:

x \in \emptyset