4x-2x^2=-32/3 — step-by-step solution

1) Let's solve in

ℝ

4 x - 2 x^{2} = - \frac{32}{3}

Since

-2 \neq 0

, divide both sides by

-2

x^{2} - 2 x = \frac{16}{3}

x^{2} - 2 x - \frac{16}{3} = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -2

and

c = - \frac{16}{3}

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot (- \frac{16}{3})

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

{(-2)}^{2}

(-2) \cdot (-2)

4

Δ = 4 - 4 \cdot (- \frac{16}{3})

ⓘ Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (- \frac{16}{3})

4 \cdot \frac{16}{3}

Δ = 4 + 4 \cdot \frac{16}{3}

ⓘ Multiplying a fraction by a number multiplies its numerator:

4 \cdot \frac{16}{3}

\frac{4 \cdot 16}{3}

\frac{64}{3}

Δ = 4 + \frac{64}{3}

ⓘ To add fractions, put them over the same denominator, then add the numerators:

4 + \frac{64}{3}

\frac{3 \cdot 4 + 64}{3}

Δ = \frac{3 \cdot 4 + 64}{3}

ⓘ Multiplications are done before additions and subtractions:

3 \cdot 4

12

Δ = \frac{12 + 64}{3}

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

12 + 64

76

Δ = \frac{76}{3}

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-2) - \sqrt{\frac{76}{3}}}{2 \cdot 1}

x_{1} = \frac{2 - \sqrt{\frac{76}{3}}}{2}

ⓘ The root of a quotient is the quotient of the roots, sqrt(a/b) = sqrt(a)/sqrt(b):

\sqrt{\frac{76}{3}}

\frac{\sqrt{76}}{\sqrt{3}}

x_{1} = \frac{2 - \frac{\sqrt{76}}{\sqrt{3}}}{2}

ⓘ We look for the largest square dividing

76

4

2^{2}

, and take it out of the root:

\sqrt{76}

\sqrt{4 \cdot 19}

\sqrt{4} \cdot \sqrt{19}

2 \cdot \sqrt{19}

■To add fractions, put them over the same denominator, then add the numerators:

2 - \frac{2 \cdot \sqrt{19}}{\sqrt{3}}

\frac{2 \cdot \sqrt{3} - 2 \cdot \sqrt{19}}{\sqrt{3}}

x_{1} = \frac{\frac{2 \cdot \sqrt{3} - 2 \cdot \sqrt{19}}{\sqrt{3}}}{2}

ⓘ Dividing a fraction by

2

is multiplying its denominator by

2

\frac{\frac{2 \cdot \sqrt{3} - 2 \cdot \sqrt{19}}{\sqrt{3}}}{2}

\frac{2 \cdot \sqrt{3} - 2 \cdot \sqrt{19}}{2 \cdot \sqrt{3}}

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

\frac{2 \cdot \sqrt{3} - 2 \cdot \sqrt{19}}{2 \cdot \sqrt{3}}

\frac{\sqrt{3} - \sqrt{19}}{\sqrt{3}}

x_{1} = \frac{\sqrt{3} - \sqrt{19}}{\sqrt{3}}

The conventional preference is to have numerical square roots in the numerator rather than in the denominator:

ⅰ

\frac{\sqrt{3} - \sqrt{19}}{\sqrt{3}}

\frac{\sqrt{3} \cdot (\sqrt{3} - \sqrt{19})}{3}

x_{1} = \frac{\sqrt{3} \cdot (\sqrt{3} - \sqrt{19})}{3}

ⓘ Expand by multiplying each term inside the parentheses:

\sqrt{3} \cdot (\sqrt{3} - \sqrt{19})

3 + \sqrt{3} \cdot (- \sqrt{19})

x_{1} = \frac{3 + \sqrt{3} \cdot (- \sqrt{19})}{3}

ⓘ The minus sign is placed in front of the product:

\sqrt{3} \cdot (- \sqrt{19})

- \sqrt{3} \cdot \sqrt{19}

x_{1} = \frac{3 - \sqrt{3} \cdot \sqrt{19}}{3}

ⓘ The product of two square roots is the square root of the product:

\sqrt{3} \cdot \sqrt{19}

\sqrt{3 \cdot 19}

x_{1} = \frac{3 - \sqrt{3 \cdot 19}}{3}

ⅰ

3 \cdot 19

3 \cdot (10 + 9)

3 \cdot 10 + 3 \cdot 9

30 + 27

57

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

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-2) + \sqrt{\frac{76}{3}}}{2 \cdot 1}

x_{2} = \frac{2 + \sqrt{\frac{76}{3}}}{2}

ⓘ The root of a quotient is the quotient of the roots, sqrt(a/b) = sqrt(a)/sqrt(b):

\sqrt{\frac{76}{3}}

\frac{\sqrt{76}}{\sqrt{3}}

x_{2} = \frac{2 + \frac{\sqrt{76}}{\sqrt{3}}}{2}

ⓘ We look for the largest square dividing

76

4

2^{2}

, and take it out of the root:

\sqrt{76}

\sqrt{4 \cdot 19}

\sqrt{4} \cdot \sqrt{19}

2 \cdot \sqrt{19}

■To add fractions, put them over the same denominator, then add the numerators:

2 + \frac{2 \cdot \sqrt{19}}{\sqrt{3}}

\frac{2 \cdot \sqrt{3} + 2 \cdot \sqrt{19}}{\sqrt{3}}

x_{2} = \frac{\frac{2 \cdot \sqrt{3} + 2 \cdot \sqrt{19}}{\sqrt{3}}}{2}

ⓘ Dividing a fraction by

2

is multiplying its denominator by

2

\frac{\frac{2 \cdot \sqrt{3} + 2 \cdot \sqrt{19}}{\sqrt{3}}}{2}

\frac{2 \cdot \sqrt{3} + 2 \cdot \sqrt{19}}{2 \cdot \sqrt{3}}

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

\frac{2 \cdot \sqrt{3} + 2 \cdot \sqrt{19}}{2 \cdot \sqrt{3}}

\frac{\sqrt{3} + \sqrt{19}}{\sqrt{3}}

x_{2} = \frac{\sqrt{3} + \sqrt{19}}{\sqrt{3}}

By convention, we prefer numerical square roots in the numerator:

ⅰ

\frac{\sqrt{3} + \sqrt{19}}{\sqrt{3}}

\frac{\sqrt{3} \cdot (\sqrt{3} + \sqrt{19})}{3}

x_{2} = \frac{\sqrt{3} \cdot (\sqrt{3} + \sqrt{19})}{3}

ⓘ Expand by multiplying each term inside the parentheses:

\sqrt{3} \cdot (\sqrt{3} + \sqrt{19})

3 + \sqrt{3} \cdot \sqrt{19}

x_{2} = \frac{3 + \sqrt{3} \cdot \sqrt{19}}{3}

ⓘ Powers with the same exponent are multiplied by multiplying the bases:

\sqrt{3} \cdot \sqrt{19}

\sqrt{3 \cdot 19}

x_{2} = \frac{3 + \sqrt{3 \cdot 19}}{3}

ⅰ

3 \cdot 19

3 \cdot (10 + 9)

3 \cdot 10 + 3 \cdot 9

30 + 27

57

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

▶Solutions obtained:

x = {\frac{3 - \sqrt{57}}{3}, \frac{3 + \sqrt{57}}{3}}

2) The numeric value of:

x = {\frac{3 - \sqrt{57}}{3}, \frac{3 + \sqrt{57}}{3}}

equals:

x = {\approx \frac{3 - 7.5498}{3}, \approx \frac{3 + 7.5498}{3}}

x = {\approx - \frac{4.5498}{3}, \approx \frac{10.5498}{3}}

▷

x = {\approx -1.5166, \approx 3.5166}