What are the solutions to the equation ((x+2))/((4))=(((x+1)(x+2)))/((2)) ?

The solutions to the equation ((x+2))/((4))=(((x+1)(x+2)))/((2)) are x=-1/2 ,x=-2

Find the zeros of ((x+2))/((4))=(((x+1)(x+2)))/((2))

The zeros of ((x+2))/((4))=(((x+1)(x+2)))/((2)) are x=-1/2 ,x=-2

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\frac{( x + 2 )}{( 4 )} = \frac{( ( x + 1 ) ( x + 2 ) )}{( 2 )}

x = - \frac{1}{2}, x = - 2

Decimal Notation

x = - 0.5, x = - 2

Solution steps

\frac{( x + 2 )}{( 4 )} = \frac{( ( x + 1 ) ( x + 2 ) )}{( 2 )}

Simplify

(4) : 4

Simplify

(2) : 2

\frac{( x + 2 )}{4} = \frac{( ( x + 1 ) ( x + 2 ) )}{2}

Find Least Common Multiplier of

4, 2 : 4

Multiply by LCM=

4

\frac{x + 2}{4} \cdot 4 = \frac{( x + 1 ) ( x + 2 )}{2} \cdot 4

Simplify

x + 2 = 2 (x + 1) (x + 2)

Expand

2 (x + 1) (x + 2) : 2 x^{2} + 6 x + 4

x + 2 = 2 x^{2} + 6 x + 4

Switch sides

2 x^{2} + 6 x + 4 = x + 2

Move

2

to the left side

2 x^{2} + 6 x + 2 = x

Move

x

to the left side

2 x^{2} + 5 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 2 = 3

x_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 3}{2 \cdot 2}, x_{2} = \frac{- 5 - 3}{2 \cdot 2}

x = \frac{- 5 + 3}{2 \cdot 2} : - \frac{1}{2}

x = \frac{- 5 - 3}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

x = - \frac{1}{2}, x = - 2

- 0.1 q^{2} + 90 = 0.2 q^{2} + q + 50

(3 x - 5)^{2} = 0

2 x^{2} - 13 x - 6 = 0

x^{2} - 8 x = - 6

x^{2} - 8 x = - 4

What are the solutions to the equation ((x+2))/((4))=(((x+1)(x+2)))/((2)) ?
The solutions to the equation ((x+2))/((4))=(((x+1)(x+2)))/((2)) are x=-1/2 ,x=-2
Find the zeros of ((x+2))/((4))=(((x+1)(x+2)))/((2))
The zeros of ((x+2))/((4))=(((x+1)(x+2)))/((2)) are x=-1/2 ,x=-2