-2/x >-(5x)/(x^2+6) — step-by-step solution

1) Let's solve in

ℝ

- \frac{2}{x} > - \frac{5 x}{x^{2} + 6}

2) Let's add fractions:

- \frac{2}{x} + \frac{5 x}{x^{2} + 6} > 0

ⓘ To add fractions, put them over the same denominator, then add the numerators:

- \frac{2}{x} + \frac{5 x}{x^{2} + 6}

\frac{- 2 \cdot (x^{2} + 6) + 5 x \cdot x}{x \cdot (x^{2} + 6)}

\frac{- 2 \cdot (x^{2} + 6) + 5 x \cdot x}{x \cdot (x^{2} + 6)} > 0

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

\frac{- 2 \cdot (x^{2} + 6) + 5 x^{2}}{x \cdot (x^{2} + 6)} > 0

We study the sign of the numerator

⇥2.1) Let's study the sign of

- 2 \cdot (x^{2} + 6) + 5 x^{2}

by solving in

ℝ

- 2 \cdot (x^{2} + 6) + 5 x^{2} > 0

⇥ⓘ Expand by multiplying each term inside the parentheses:

- 2 \cdot (x^{2} + 6)

- (2 x^{2} + 12)

⇥

- (2 x^{2} + 12) + 5 x^{2} > 0

⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (2 x^{2} + 12)

- 2 x^{2} - 12

⇥

- 2 x^{2} - 12 + 5 x^{2} > 0

⇥ⓘ Like terms are grouped by adding their coefficients:

- 2 x^{2} + 5 x^{2}

3 x^{2}

⇥

3 x^{2} - 12 > 0

⇥Since

3 > 0

, divide both sides by

3

without reversing the inequality:

x^{2} - 4 > 0

⇥An even power only depends on the absolute value:

x^{2} > 4

is equivalent to

| x | > 2

⇥

| A | > B

is equivalent to:

A > B

A < - B

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

⇥2.2) Let's solve in

ℝ

x \cdot (x^{2} + 6) > 0

⇥We study the sign of each factor of the product

⇥⇥The expression

x

is zero at

x = 0

(forbidden value), it is negative before and positive after

⇥Let's study the sign of

x^{2} + 6

by solving in

ℝ

x^{2} + 6 > 0

⇥In ℝ, an even power is never negative:

x^{2} > -6

is true for every

x

⇥The simplified relation is always true; the original expression's domain must still be respected

⇥⇥So the expression

x^{2} + 6

is always positive

3) Let's draw up the sign table:

Candidates, pending the domain check:

x \in (-2, 0) \cup (2, +∞)

4) Let's evaluate the domain of

- \frac{2}{x} > - \frac{5 x}{x^{2} + 6}

For the expression

- \frac{2}{x}

to exist, you must check:

x \neq 0

For the expression

- \frac{5 x}{x^{2} + 6}

to be defined, we must have:

x^{2} + 6 \neq 0

The condition

x^{2} + 6 \neq 0

is always true on its real domain

Domain constraints:

x \neq 0

▶Solution:

x \in (-2, 0) \cup (2, +∞)