x^2-12<-x — step-by-step solution

Let's solve in

ℝ

x^{2} - 12 < - x

x^{2} + 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 = 1

and

c = -12

Δ = 1 - 4 \cdot 1 \cdot (-12)

Δ = 1 - 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 = 1 + 48

ⓘ Start from

1

and move forward

48

on the number line:

1 + 48

49

Δ = 49

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 1 - \sqrt{49}}{2 \cdot 1}

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

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

- 1 - 7

-8

x_{1} = - \frac{8}{2}

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

2

, their greatest common divisor:

- \frac{8}{2}

-4

x_{1} = -4

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 1 + \sqrt{49}}{2 \cdot 1}

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

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

- 1 + 7

6

x_{2} = \frac{6}{2}

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

2

, their greatest common divisor:

\frac{6}{2}

3

x_{2} = 3

The quadratic has the opposite sign to a =

1

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

▶Solution obtained:

x \in (-4, 3)