domain of f(x)=(x+2)/(x^2-3x-10)

We save the definition:

f (x) = \frac{x + 2}{x^{2} - 3 x - 10}

Let's evaluate the domain of

\frac{x + 2}{x^{2} - 3 x - 10}

For the expression

\frac{x + 2}{x^{2} - 3 x - 10}

to exist, you must check:

x^{2} - 3 x - 10 \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 = -3

and

c = -10

Δ = {(-3)}^{2} - 4 \cdot 1 \cdot (-10)

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

{(-3)}^{2}

(-3) \cdot (-3)

9

Δ = 9 - 4 \cdot (-10)

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

- 4 \cdot (-10)

40

Δ = 9 + 40

ⓘ Start from

9

and move forward

40

on the number line:

9 + 40

49

Δ = 49

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-3) - \sqrt{49}}{2 \cdot 1}

x_{1} = \frac{3 - \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{3 - 7}{2}

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

3 - 7

-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{- (-3) + \sqrt{49}}{2 \cdot 1}

x_{2} = \frac{3 + \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{3 + 7}{2}

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

3 + 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

▶Domain of definition:

x \in ℝ \ {-2, 5}