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

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

Find the zeros of (x^2-x)/6-(x^2+x)/3 =0

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

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

x = - 3, x = 0

Solution steps

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

Find Least Common Multiplier of

6, 3 : 6

Multiply by LCM=

6

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

Simplify

x^{2} - x - 2 (x^{2} + x) = 0

Expand

x^{2} - x - 2 (x^{2} + x) : - x^{2} - 3 x

- x^{2} - 3 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- ( - 3 ) - 3}{2 ( - 1 )} : 0

The solutions to the quadratic equation are:

x = - 3, x = 0

∣ x + 5 ∣ > - 2

x^{2} - 16 x + 31 = 0

- 2.5 \leq \frac{1 - 3 y}{2} \leq 1.5

11 a = - 16

5 a^{2} + 60 a + 100

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