(x^2-x-2)/(x^2-4x+3)>= 0 — step-by-step solution

1) Let's solve in

ℝ

\frac{x^{2} - x - 2}{x^{2} - 4 x + 3} \geq 0

We study the sign of the numerator

⇥1.1) Let's study the sign of

x^{2} - x - 2

by solving in

ℝ

x^{2} - x - 2 \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 = -1

and

c = -2

⇥

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-2)

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

{(-1)}^{2}

(-1) \cdot (-1)

1

⇥

Δ = 1 - 4 \cdot (-2)

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

- 4 \cdot (-2)

8

⇥

Δ = 1 + 8

⇥ⓘ Start from

1

and move forward

8

on the number line:

1 + 8

9

⇥

Δ = 9

⇥Since

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{9}}{2 \cdot 1}

⇥

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

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

1 - 3

-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{- (-1) + \sqrt{9}}{2 \cdot 1}

⇥

x_{2} = \frac{1 + \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{1 + 3}{2}

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

1 + 3

4

⇥

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

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

2

, their greatest common divisor:

\frac{4}{2}

2

⇥

x_{2} = 2

⇥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.

⇥Candidates, pending the domain check:

x \in (-∞, -1] \cup [2, +∞)

⇥So the expression

x^{2} - x - 2

is positive or zero on

(-∞, -1] \cup [2, +∞)

and negative otherwise

We study the sign of the denominator, which must not be zero

⇥1.2) Let's study the sign of

x^{2} - 4 x + 3

by solving in

ℝ

x^{2} - 4 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 = -4

and

c = 3

⇥

Δ = {(-4)}^{2} - 4 \cdot 1 \cdot 3

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

{(-4)}^{2}

(-4) \cdot (-4)

16

⇥

Δ = 16 - 4 \cdot 3

⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 3

12

⇥

Δ = 16 - 12

⇥ⓘ Start from

16

and move back

12

on the number line:

16 - 12

4

⇥

Δ = 4

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-4) - \sqrt{4}}{2 \cdot 1}

⇥

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

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

\sqrt{4}

2

;

2^{2}

4

⇥

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

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

4 - 2

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{- (-4) + \sqrt{4}}{2 \cdot 1}

⇥

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

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

\sqrt{4}

2

;

2^{2}

4

⇥

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

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

4 + 2

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.

⇥Candidates, pending the domain check:

x \in (-∞, 1) \cup (3, +∞)

⇥So the expression

x^{2} - 4 x + 3

is positive on

(-∞, 1) \cup (3, +∞)

and negative or zero otherwise

2) Let's draw up the sign table:

x	$-∞$		$-1$		$1$		$2$		$3$		$+∞$
$x^{2} - x - 2$		+	0	-		-	0	+		+
$x^{2} - 4 x + 3$		+		+	0	-		-	0	+
$\frac{x^{2} - x - 2}{x^{2} - 4 x + 3}$		+	0	-	❌	+	0	-	❌	+

Candidates, pending the domain check:

x \in (-∞, -1] \cup (1, 2] \cup (3, +∞)

3) Let's find the domain of

\frac{x^{2} - x - 2}{x^{2} - 4 x + 3} \geq 0

For the expression

\frac{x^{2} - x - 2}{x^{2} - 4 x + 3}

to exist, you must check:

x^{2} - 4 x + 3 \neq 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 = -4

and

c = 3

Δ = {(-4)}^{2} - 4 \cdot 1 \cdot 3

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

{(-4)}^{2}

(-4) \cdot (-4)

16

Δ = 16 - 4 \cdot 3

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 3

12

Δ = 16 - 12

ⓘ Start from

16

and move back

12

on the number line:

16 - 12

4

Δ = 4

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-4) - \sqrt{4}}{2 \cdot 1}

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

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

\sqrt{4}

2

;

2^{2}

4

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

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

4 - 2

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{- (-4) + \sqrt{4}}{2 \cdot 1}

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

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

\sqrt{4}

2

;

2^{2}

4

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

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

4 + 2

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

Domain constraints:

x \in ℝ \ {1, 3}

▶Solution:

x \in (-∞, -1] \cup (1, 2] \cup (3, +∞)