domain of (2x^2+3x-2)/(x^2+x-2) — step-by-step solution

Let's find the domain of

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

For the expression

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

to exist, you must check:

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

and

c = -2

Δ = 1 - 4 \cdot 1 \cdot (-2)

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

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

-4

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

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

2

x_{2} = \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_{2} = 1

▶Domain of definition:

x \in ℝ \ {-2, 1}