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

The solutions to the equation (x^2+2)/3 =(2x)/4 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4

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

The zeros of (x^2+2)/3 =(2x)/4 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4

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

x = \frac{3}{4} + i \frac{23}{4}, x = \frac{3}{4} - i \frac{23}{4}

Solution steps

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

Find Least Common Multiplier of

3, 4 : 12

Multiply by LCM=

12

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

Simplify

4 (x^{2} + 2) = 6 x

Expand

4 (x^{2} + 2) : 4 x^{2} + 8

4 x^{2} + 8 = 6 x

Move

6 x

to the left side

4 x^{2} + 8 - 6 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} - 6 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 \cdot 8}{2 \cdot 4}

Simplify

(- 6)^{2} - 4 \cdot 4 \cdot 8 : 223 i

x_{1, 2} = \frac{- ( - 6 ) \pm 2 23 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 23 i}{2 \cdot 4}, x_{2} = \frac{- ( - 6 ) - 2 23 i}{2 \cdot 4}

x = \frac{- ( - 6 ) + 2 23 i}{2 \cdot 4} : \frac{3}{4} + i \frac{23}{4}

x = \frac{- ( - 6 ) - 2 23 i}{2 \cdot 4} : \frac{3}{4} - i \frac{23}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{23}{4}, x = \frac{3}{4} - i \frac{23}{4}

25 x^{2} = 60 x - 37

x^{2} - 9 = 55

(x - 1)^{2} + ((\frac{- 5}{12}) (\frac{x + 8}{5}))^{2} = + 1

x^{2} + 2 = 3 x

x, z = 4 - x^{2}

What are the solutions to the equation (x^2+2)/3 =(2x)/4 ?
The solutions to the equation (x^2+2)/3 =(2x)/4 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4
Find the zeros of (x^2+2)/3 =(2x)/4
The zeros of (x^2+2)/3 =(2x)/4 are x= 3/4+i(sqrt(23))/4 ,x= 3/4-i(sqrt(23))/4