What are the solutions to the equation (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4 ?

The solutions to the equation (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4 are x=-2/3 ,x=4

Find the zeros of (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4

The zeros of (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4 are x=-2/3 ,x=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 2) x - \frac{x + 2}{3} - \frac{( x - 2 ) ( x + 2 )}{2} = (x - 2)^{2} - 4

x = - \frac{2}{3}, x = 4

Decimal Notation

x = - 0.66666 \dots, x = 4

Solution steps

(x - 2) x - \frac{x + 2}{3} - \frac{( x - 2 ) ( x + 2 )}{2} = (x - 2)^{2} - 4

Find Least Common Multiplier of

3, 2 : 6

Multiply by LCM=

6

(x - 2) x \cdot 6 - \frac{x + 2}{3} \cdot 6 - \frac{( x - 2 ) ( x + 2 )}{2} \cdot 6 = (x - 2)^{2} \cdot 6 - 4 \cdot 6

Simplify

6 x (x - 2) - 2 (x + 2) - 3 (x - 2) (x + 2) = 6 (x - 2)^{2} - 24

Expand

6 x (x - 2) - 2 (x + 2) - 3 (x - 2) (x + 2) : 3 x^{2} - 14 x + 8

Expand

6 (x - 2)^{2} - 24 : 6 x^{2} - 24 x

3 x^{2} - 14 x + 8 = 6 x^{2} - 24 x

Move

24 x

to the left side

3 x^{2} + 10 x + 8 = 6 x^{2}

Move

6 x^{2}

to the left side

- 3 x^{2} + 10 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 3 ) \cdot 8}{2 ( - 3 )}

1 0^{2} - 4 (- 3) \cdot 8 = 14

x_{1, 2} = \frac{- 10 \pm 14}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 10 + 14}{2 ( - 3 )}, x_{2} = \frac{- 10 - 14}{2 ( - 3 )}

x = \frac{- 10 + 14}{2 ( - 3 )} : - \frac{2}{3}

x = \frac{- 10 - 14}{2 ( - 3 )} : 4

The solutions to the quadratic equation are:

x = - \frac{2}{3}, x = 4

7 x + 8 - 9 (x + 1) = 3 x + 2

\frac{x ^{\frac{2}{5}} y}{x y ^{- \frac{1}{3}}}

x^{2} + \frac{8 x}{3} = - \frac{7}{9}

\frac{1}{x + 3} > (x + 1)

4 ∣ 3 a + 4 ∣ = 16 a + 8

What are the solutions to the equation (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4 ?
The solutions to the equation (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4 are x=-2/3 ,x=4
Find the zeros of (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4
The zeros of (x-2)x-(x+2)/3-((x-2)(x+2))/2 =(x-2)^2-4 are x=-2/3 ,x=4