-3*(x+1)^2-12<0 — step-by-step solution

Let's solve in

ℝ

- 3 \cdot {(x + 1)}^{2} - 12 < 0

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

{(x + 1)}^{2}

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

- 3 \cdot (x + 1) \cdot (x + 1) - 12 < 0

ⓘ Expand by multiplying each term inside the parentheses:

3 \cdot (x + 1)

3 x + 3

- (3 x + 3) \cdot (x + 1) - 12 < 0

ⓘ Expand by multiplying each term inside the parentheses:

- (3 x + 3) \cdot (x + 1)

- (3 \cdot x \cdot x

+ 3 x \cdot 1

+ 3 x

+ 3 \cdot 1)

- (3 \cdot x \cdot x

+ 3 x \cdot 1

+ 3 x

+ 3 \cdot 1) - 12 < 0

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

- (3 \cdot x \cdot x

+ 3 x \cdot 1

+ 3 x

+ 3 \cdot 1)

- 3 \cdot x \cdot x

- 3 x \cdot 1

- 3 x

- 3 \cdot 1

- 3 \cdot x \cdot x

- 3 x \cdot 1

- 3 x

- 3 \cdot 1

- 12 < 0

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

- 3 x^{2}

- 3 x

- 3 x

- 3

- 12 < 0

ⓘ Like terms are grouped by adding their coefficients:

- 3 x - 3 x

-6 x

■Start from

-3

and move back

12

more on the number line, away from zero:

- 3 - 12

-15

- 3 x^{2} - 6 x - 15 < 0

Since

-3 < 0

, divide both sides by

-3

and reverse the inequality:

x^{2} + 2 x + 5 > 0

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 2

and

c = 5

Δ = 2^{2} - 4 \cdot 1 \cdot 5

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

2^{2}

2 \cdot 2

4

Δ = 4 - 4 \cdot 5

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 5

20

Δ = 4 - 20

ⓘ Start from

4

and move back

20

on the number line, going below zero:

4 - 20

-16

Δ = -16

The quadratic has no real root: it has the sign of a =

1

everywhere, so it is indeed positive for every

x

▶Solution found:

x \in ℝ