What are the solutions to the equation (x^2)/2 = 1/6+x/(12) ?

The solutions to the equation (x^2)/2 = 1/6+x/(12) are x= 2/3 ,x=-1/2

Find the zeros of (x^2)/2 = 1/6+x/(12)

The zeros of (x^2)/2 = 1/6+x/(12) are x= 2/3 ,x=-1/2

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\frac{x ^{2}}{2} = \frac{1}{6} + \frac{x}{12}

x = \frac{2}{3}, x = - \frac{1}{2}

Decimal Notation

x = 0.66666 \dots, x = - 0.5

Solution steps

\frac{x ^{2}}{2} = \frac{1}{6} + \frac{x}{12}

Find Least Common Multiplier of

2, 6, 12 : 12

Multiply by LCM=

12

\frac{x ^{2}}{2} \cdot 12 = \frac{1}{6} \cdot 12 + \frac{x}{12} \cdot 12

Simplify

6 x^{2} = 2 + x

Move

x

to the left side

6 x^{2} - x = 2

Move

2

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 ( - 2 )}{2 \cdot 6}

(- 1)^{2} - 4 \cdot 6 (- 2) = 7

x_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 6}, x_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 6}

x = \frac{- ( - 1 ) + 7}{2 \cdot 6} : \frac{2}{3}

x = \frac{- ( - 1 ) - 7}{2 \cdot 6} : - \frac{1}{2}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = - \frac{1}{2}

x^{2} - 2 x + \frac{1}{2} = 0

4 x^{2} - 2 x + 2 = 0

x^{2} - 16 x - 192 = 0

3 y^{2} - 6 y + 0 = 0

- 5 x^{2} - 6 x = - 1 + 2 x

What are the solutions to the equation (x^2)/2 = 1/6+x/(12) ?
The solutions to the equation (x^2)/2 = 1/6+x/(12) are x= 2/3 ,x=-1/2
Find the zeros of (x^2)/2 = 1/6+x/(12)
The zeros of (x^2)/2 = 1/6+x/(12) are x= 2/3 ,x=-1/2