-36x(-9x^2+25)>0 — step-by-step solution

1) Let's solve in

ℝ

-36 x \cdot (- 9 x^{2} + 25) > 0

We study the sign of each factor of the product

⇥The expression

- 36 x

is zero at:

⇥

x = 0

⇥Its coefficient of

x

-36

, being negative, it is positive before and negative after

0

⇥1.1) Let's study the sign of

- 9 x^{2} + 25

by widening the sign to ≥ and solving in

ℝ

- 9 x^{2} + 25 \geq 0

⇥For

- 9 x^{2} + 25

, we recognize a difference of two squares:

5^{2} - {(3 x)}^{2}

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = 5

and

b = 3 x

⇥You can write the expression:

- 9 x^{2} + 25

in factored form:

(5 - 3 x) \cdot (5 + 3 x)

⇥1.2) Let's solve in

ℝ

(5 - 3 x) \cdot (5 + 3 x) \geq 0

⇥We study the sign of each factor of the product

⇥⇥The expression

5 - 3 x

is zero at:

⇥⇥

x = \frac{-5}{-3}

⇥⇥ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{-5}{-3}

\frac{5}{3}

⇥⇥

x = \frac{5}{3}

⇥⇥Its coefficient of

x

-3

, being negative, it is positive before and negative after

\frac{5}{3}

⇥⇥The expression

5 + 3 x

is zero at:

⇥⇥

x = - \frac{5}{3}

⇥⇥Its coefficient of

x

3

, being positive, it is negative before and positive after

- \frac{5}{3}

2) Let's draw up the sign table:

▶Solution found:

x \in (- \frac{5}{3}, 0) \cup (\frac{5}{3}, +∞)

3) The numeric value of:

x \in (- \frac{5}{3}, 0) \cup (\frac{5}{3}, +∞)

is:

▷

x \in (\approx -1.6667, 0) \cup (\approx 1.6667, +∞)