x^2-2x-3>0 — step-by-step solution

Let's solve in

ℝ

x^{2} - 2 x - 3 > 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 = -2

and

c = -3

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot (-3)

ⓘ 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 (-3)

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

- 4 \cdot (-3)

12

Δ = 4 + 12

ⓘ Start from

4

and move forward

12

on the number line:

4 + 12

16

Δ = 16

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-2) - \sqrt{16}}{2 \cdot 1}

x_{1} = \frac{2 - \sqrt{16}}{2}

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

\sqrt{16}

4

;

4^{2}

16

x_{1} = \frac{2 - 4}{2}

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

2 - 4

-2

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

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

2

, their greatest common divisor:

- \frac{2}{2}

-1

x_{1} = -1

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-2) + \sqrt{16}}{2 \cdot 1}

x_{2} = \frac{2 + \sqrt{16}}{2}

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

\sqrt{16}

4

;

4^{2}

16

x_{2} = \frac{2 + 4}{2}

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

2 + 4

6

x_{2} = \frac{6}{2}

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

2

, their greatest common divisor:

\frac{6}{2}

3

x_{2} = 3

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 excluded.

▶Solution obtained:

x \in (-∞, -1) \cup (3, +∞)