What are the solutions to the equation 1/6 x^2+1/2 x+1/3 =0 ?

The solutions to the equation 1/6 x^2+1/2 x+1/3 =0 are x=-1,x=-2

Find the zeros of 1/6 x^2+1/2 x+1/3 =0

The zeros of 1/6 x^2+1/2 x+1/3 =0 are x=-1,x=-2

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\frac{1}{6} x^{2} + \frac{1}{2} x + \frac{1}{3} = 0

x = - 1, x = - 2

Solution steps

\frac{1}{6} x^{2} + \frac{1}{2} x + \frac{1}{3} = 0

Find Least Common Multiplier of

6, 2, 3 : 6

Multiply by LCM=

6

\frac{1}{6} x^{2} \cdot 6 + \frac{1}{2} x \cdot 6 + \frac{1}{3} \cdot 6 = 0 \cdot 6

Simplify

x^{2} + 3 x + 2 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot 2 = 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = - 2

3 (x + 2) - 10 = 4 x - 6 + x

∣ 2 x - 7 ∣ < - 3

- 1 + \frac{7}{4}

2 [x - (4 + 2 x) + 3] = 2 x + 2

36 + 12 x + x^{2}

What are the solutions to the equation 1/6 x^2+1/2 x+1/3 =0 ?
The solutions to the equation 1/6 x^2+1/2 x+1/3 =0 are x=-1,x=-2
Find the zeros of 1/6 x^2+1/2 x+1/3 =0
The zeros of 1/6 x^2+1/2 x+1/3 =0 are x=-1,x=-2