x^2+21x-20>= 4x-80 — step-by-step solution

Let's solve in

ℝ

x^{2} + 21 x - 20 \geq 4 x - 80

x^{2}

+ 21 x

- 20

- 4 x

+ 80 \geq 0

ⓘ Start from

-20

and move forward

80

on the number line, going past zero:

- 20 + 80

60

■Like terms are grouped by adding their coefficients:

21 x - 4 x

17 x

x^{2} + 17 x + 60 \geq 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 = 17

and

c = 60

Δ = 17^{2} - 4 \cdot 1 \cdot 60

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

17^{2}

17 \cdot 17

289

Δ = 289 - 4 \cdot 60

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 60

240

Δ = 289 - 240

ⓘ Start from

289

and move back

240

on the number line:

289 - 240

49

Δ = 49

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 17 - \sqrt{49}}{2 \cdot 1}

x_{1} = \frac{- 17 - \sqrt{49}}{2}

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

\sqrt{49}

7

;

7^{2}

49

x_{1} = \frac{- 17 - 7}{2}

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

- 17 - 7

-24

x_{1} = - \frac{24}{2}

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

2

, their greatest common divisor:

- \frac{24}{2}

-12

x_{1} = -12

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 17 + \sqrt{49}}{2 \cdot 1}

x_{2} = \frac{- 17 + \sqrt{49}}{2}

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

\sqrt{49}

7

;

7^{2}

49

x_{2} = \frac{- 17 + 7}{2}

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

- 17 + 7

-10

x_{2} = - \frac{10}{2}

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

2

, their greatest common divisor:

- \frac{10}{2}

-5

x_{2} = -5

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 obtained:

x \in (-∞, -12] \cup [-5, +∞)