4x^2+9x<-5 — step-by-step solution

1) Let's solve in

ℝ

4 x^{2} + 9 x < -5

4 x^{2} + 9 x + 5 < 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 4

b = 9

and

c = 5

Δ = 9^{2} - 4 \cdot 4 \cdot 5

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

9^{2}

9 \cdot 9

81

■Multiplications are done before additions and subtractions:

4 \cdot 4

16

Δ = 81 - 16 \cdot 5

ⓘ Multiplications are done before additions and subtractions:

16 \cdot 5

(10 + 6) \cdot 5

10 \cdot 5 + 6 \cdot 5

50 + 30

80

Δ = 81 - 80

ⓘ Start from

81

and move back

80

on the number line:

81 - 80

1

Δ = 1

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 9 - \sqrt{1}}{2 \cdot 4}

x_{1} = \frac{- 9 - 1}{2 \cdot 4}

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

- 9 - 1

-10

ⅰ

2 \cdot 4

8

x_{1} = - \frac{10}{8}

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

2

, their greatest common divisor:

- \frac{10}{8}

- \frac{5 \cdot 2}{4 \cdot 2}

- \frac{5}{4}

x_{1} = - \frac{5}{4}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 9 + \sqrt{1}}{2 \cdot 4}

x_{2} = \frac{- 9 + 1}{2 \cdot 4}

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

- 9 + 1

-8

ⅰ

2 \cdot 4

8

x_{2} = - \frac{8}{8}

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

8

, their greatest common divisor:

- \frac{8}{8}

-1

x_{2} = -1

The quadratic has the opposite sign to a =

4

between its roots, so it is indeed negative between its roots; the roots, where it is zero, are excluded.

▶Solution obtained:

x \in (- \frac{5}{4}, -1)

2) The numeric value of:

x \in (- \frac{5}{4}, -1)

equals:

▷

x \in (-1.25, -1)