x^2+11x+28>= 0 — step-by-step solution

Let's solve in

ℝ

x^{2} + 11 x + 28 \geq 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 = 11

and

c = 28

Δ = 11^{2} - 4 \cdot 1 \cdot 28

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

11^{2}

11 \cdot 11

121

Δ = 121 - 4 \cdot 28

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 28

4 \cdot (20 + 8)

4 \cdot 20 + 4 \cdot 8

80 + 32

112

Δ = 121 - 112

ⓘ Start from

121

and move back

112

on the number line:

121 - 112

9

Δ = 9

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 11 - \sqrt{9}}{2 \cdot 1}

x_{1} = \frac{- 11 - \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{1} = \frac{- 11 - 3}{2}

ⓘ The numerator and the denominator are computed before the division:

- 11 - 3

-14

x_{1} = - \frac{14}{2}

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

2

, their greatest common divisor:

- \frac{14}{2}

-7

x_{1} = -7

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 11 + \sqrt{9}}{2 \cdot 1}

x_{2} = \frac{- 11 + \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{2} = \frac{- 11 + 3}{2}

ⓘ The numerator and the denominator are computed before the division:

- 11 + 3

-8

x_{2} = - \frac{8}{2}

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

2

, their greatest common divisor:

- \frac{8}{2}

-4

x_{2} = -4

The quadratic has the sign of a =

1

outside its roots, so it is indeed positive outside its roots; the roots, where it is zero, are included.

▶Solution found:

x \in (-∞, -7] \cup [-4, +∞)