114-36k-6k^2>0 — step-by-step solution

1) Let's solve in

ℝ

114 - 36 k - 6 k^{2} > 0

Since

-6 < 0

, divide both sides by

-6

and reverse the inequality:

- 19 + 6 k + k^{2} < 0

Given the quadratic form

a k^{2} + b k + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 6

and

c = -19

Δ = 6^{2} - 4 \cdot 1 \cdot (-19)

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

6^{2}

6 \cdot 6

36

Δ = 36 - 4 \cdot (-19)

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

- 4 \cdot (-19)

4 \cdot 19

4 \cdot (10 + 9)

4 \cdot 10 + 4 \cdot 9

40 + 36

76

Δ = 36 + 76

ⓘ Start from

36

and move forward

76

on the number line:

36 + 76

112

Δ = 112

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

k_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 6 - \sqrt{112}}{2 \cdot 1}

k_{1} = \frac{- 6 - \sqrt{112}}{2}

ⓘ We look for the largest square dividing

112

16

4^{2}

, and take it out of the root:

\sqrt{112}

\sqrt{16 \cdot 7}

\sqrt{16} \cdot \sqrt{7}

4 \cdot \sqrt{7}

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

- 6 - 4 \cdot \sqrt{7}

- (6 + 4 \cdot \sqrt{7})

k_{1} = - \frac{6 + 4 \cdot \sqrt{7}}{2}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

- \frac{6 + 4 \cdot \sqrt{7}}{2}

- \frac{2 \cdot (3 + 2 \cdot \sqrt{7})}{2}

- (3 + 2 \cdot \sqrt{7})

k_{1} = - (3 + 2 \cdot \sqrt{7})

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

- (3 + 2 \cdot \sqrt{7})

- 3 - 2 \cdot \sqrt{7}

k_{1} = - 3 - 2 \cdot \sqrt{7}

The quadratic formula gives

k_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 6 + \sqrt{112}}{2 \cdot 1}

k_{2} = \frac{- 6 + \sqrt{112}}{2}

ⓘ We look for the largest square dividing

112

16

4^{2}

, and take it out of the root:

\sqrt{112}

\sqrt{16 \cdot 7}

\sqrt{16} \cdot \sqrt{7}

4 \cdot \sqrt{7}

k_{2} = \frac{- 6 + 4 \cdot \sqrt{7}}{2}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{- 6 + 4 \cdot \sqrt{7}}{2}

\frac{2 \cdot (- 3 + 2 \cdot \sqrt{7})}{2}

- 3 + 2 \cdot \sqrt{7}

k_{2} = - 3 + 2 \cdot \sqrt{7}

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:

k \in (- 2 \cdot \sqrt{7} - 3, 2 \cdot \sqrt{7} - 3)

2) The numeric value of:

k \in (- 2 \cdot \sqrt{7} - 3, 2 \cdot \sqrt{7} - 3)

equals:

k \in (- 2 \cdot \approx 2.6458 - 3, 2 \cdot \approx 2.6458 - 3)

k \in (- \approx 5.2915 - 3, \approx 5.2915 - 3)

▷

k \in (\approx -8.2915, \approx 2.2915)