x^2+3x-12>0 — step-by-step solution

1) Let's solve in

ℝ

x^{2} + 3 x - 12 > 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 = -12

Δ = 3^{2} - 4 \cdot 1 \cdot (-12)

ⓘ 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 (-12)

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

- 4 \cdot (-12)

4 \cdot 12

4 \cdot (10 + 2)

4 \cdot 10 + 4 \cdot 2

40 + 8

48

Δ = 9 + 48

ⓘ Start from

9

and move forward

48

on the number line:

9 + 48

57

Δ = 57

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 3 - \sqrt{57}}{2 \cdot 1}

x_{1} = \frac{- 3 - \sqrt{57}}{2}

ⓘ The minus sign is factored out, changing the sign of each term:

- 3 - \sqrt{57}

- (3 + \sqrt{57})

x_{1} = - \frac{3 + \sqrt{57}}{2}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 3 + \sqrt{57}}{2 \cdot 1}

x_{2} = \frac{- 3 + \sqrt{57}}{2}

The quadratic has the sign of a =

1

outside its roots, so it is indeed positive outside its roots; the roots, where it is zero, are excluded.

▶Solution obtained:

x \in (-∞, - \frac{3 + \sqrt{57}}{2}) \cup (\frac{\sqrt{57} - 3}{2}, +∞)

2) The numeric value of:

x \in (-∞, - \frac{3 + \sqrt{57}}{2}) \cup (\frac{\sqrt{57} - 3}{2}, +∞)

is:

x \in (-∞, - \frac{3 + \approx 7.5498}{2}) \cup (\frac{\approx 7.5498 - 3}{2}, +∞)

x \in (-∞, - \frac{\approx 10.5498}{2}) \cup (\frac{\approx 4.5498}{2}, +∞)

▷

x \in (-∞, \approx -5.2749) \cup (\approx 2.2749, +∞)