What are the solutions to the equation ((k+1))/2 (k+2)= 1/6 (k+3) ?

The solutions to the equation ((k+1))/2 (k+2)= 1/6 (k+3) are k=(-4+sqrt(7))/3 ,k=-(4+sqrt(7))/3

Find the zeros of ((k+1))/2 (k+2)= 1/6 (k+3)

The zeros of ((k+1))/2 (k+2)= 1/6 (k+3) are k=(-4+sqrt(7))/3 ,k=-(4+sqrt(7))/3

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\frac{( k + 1 )}{2} (k + 2) = \frac{1}{6} (k + 3)

k = \frac{- 4 + 7}{3}, k = - \frac{4 + 7}{3}

Solution steps

\frac{( k + 1 )}{2} (k + 2) = \frac{1}{6} (k + 3)

Find Least Common Multiplier of

2, 6 : 6

Multiply by LCM=

6

\frac{k + 1}{2} (k + 2) \cdot 6 = \frac{1}{6} (k + 3) \cdot 6

Simplify

3 (k + 1) (k + 2) = k + 3

Expand

3 (k + 1) (k + 2) : 3 k^{2} + 9 k + 6

3 k^{2} + 9 k + 6 = k + 3

Move

3

to the left side

3 k^{2} + 9 k + 3 = k

Move

k

to the left side

3 k^{2} + 8 k + 3 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

8^{2} - 4 \cdot 3 \cdot 3 = 27

k_{1, 2} = \frac{- 8 \pm 2 7}{2 \cdot 3}

Separate the solutions

k_{1} = \frac{- 8 + 2 7}{2 \cdot 3}, k_{2} = \frac{- 8 - 2 7}{2 \cdot 3}

k = \frac{- 8 + 2 7}{2 \cdot 3} : \frac{- 4 + 7}{3}

k = \frac{- 8 - 2 7}{2 \cdot 3} : - \frac{4 + 7}{3}

The solutions to the quadratic equation are:

k = \frac{- 4 + 7}{3}, k = - \frac{4 + 7}{3}

0.25 Q + 0.1 (2 Q) + 0.05 (2 Q + 14) = 4

\frac{x y}{x + y}

x \frac{3}{4} = 27

7 - 6

x^{2} + x^{2} = 1 2^{2}

What are the solutions to the equation ((k+1))/2 (k+2)= 1/6 (k+3) ?
The solutions to the equation ((k+1))/2 (k+2)= 1/6 (k+3) are k=(-4+sqrt(7))/3 ,k=-(4+sqrt(7))/3
Find the zeros of ((k+1))/2 (k+2)= 1/6 (k+3)
The zeros of ((k+1))/2 (k+2)= 1/6 (k+3) are k=(-4+sqrt(7))/3 ,k=-(4+sqrt(7))/3