What are the solutions to the equation x/3+(x^2)/4 =(5x)/6 ?

The solutions to the equation x/3+(x^2)/4 =(5x)/6 are x=2,x=0

Find the zeros of x/3+(x^2)/4 =(5x)/6

The zeros of x/3+(x^2)/4 =(5x)/6 are x=2,x=0

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\frac{x}{3} + \frac{x ^{2}}{4} = \frac{5 x}{6}

x = 2, x = 0

Solution steps

\frac{x}{3} + \frac{x ^{2}}{4} = \frac{5 x}{6}

Find Least Common Multiplier of

3, 4, 6 : 12

Multiply by LCM=

12

\frac{x}{3} \cdot 12 + \frac{x ^{2}}{4} \cdot 12 = \frac{5 x}{6} \cdot 12

Simplify

4 x + 3 x^{2} = 10 x

Move

10 x

to the left side

3 x^{2} - 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 6)^{2} - 4 \cdot 3 \cdot 0 = 6

x_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6}{2 \cdot 3}, x_{2} = \frac{- ( - 6 ) - 6}{2 \cdot 3}

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

x = \frac{- ( - 6 ) - 6}{2 \cdot 3} : 0

The solutions to the quadratic equation are:

x = 2, x = 0

x^{6} + 2 x^{3} - 3

3 - 2 (9 + 2 m) = m

∣ x - 1 ∣ \leq 15

\frac{10 !}{1 0 ^{k} ( 10 - k ) !}

8 1^{\frac{5}{4}}

What are the solutions to the equation x/3+(x^2)/4 =(5x)/6 ?
The solutions to the equation x/3+(x^2)/4 =(5x)/6 are x=2,x=0
Find the zeros of x/3+(x^2)/4 =(5x)/6
The zeros of x/3+(x^2)/4 =(5x)/6 are x=2,x=0