What are the solutions to the equation 4x^2=x-2 ?

The solutions to the equation 4x^2=x-2 are x= 1/8+i(sqrt(31))/8 ,x= 1/8-i(sqrt(31))/8

Find the zeros of 4x^2=x-2

The zeros of 4x^2=x-2 are x= 1/8+i(sqrt(31))/8 ,x= 1/8-i(sqrt(31))/8

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4 x^{2} = x - 2

x = \frac{1}{8} + i \frac{31}{8}, x = \frac{1}{8} - i \frac{31}{8}

Solution steps

4 x^{2} = x - 2

Move

2

to the left side

4 x^{2} + 2 = x

Move

x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 4 \cdot 2 : 31 i

x_{1, 2} = \frac{- ( - 1 ) \pm 31 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 31 i}{2 \cdot 4}, x_{2} = \frac{- ( - 1 ) - 31 i}{2 \cdot 4}

x = \frac{- ( - 1 ) + 31 i}{2 \cdot 4} : \frac{1}{8} + i \frac{31}{8}

x = \frac{- ( - 1 ) - 31 i}{2 \cdot 4} : \frac{1}{8} - i \frac{31}{8}

The solutions to the quadratic equation are:

x = \frac{1}{8} + i \frac{31}{8}, x = \frac{1}{8} - i \frac{31}{8}

5 y^{2} - 28 y + 15 = 0

24 x^{2} - 280 x + 600 = 0

x^{2} - (m^{2} - 5) x - 8 m + 3 = 0

6 + 4 x - x^{2}

\frac{x ^{2}}{3} = 27

What are the solutions to the equation 4x^2=x-2 ?
The solutions to the equation 4x^2=x-2 are x= 1/8+i(sqrt(31))/8 ,x= 1/8-i(sqrt(31))/8
Find the zeros of 4x^2=x-2
The zeros of 4x^2=x-2 are x= 1/8+i(sqrt(31))/8 ,x= 1/8-i(sqrt(31))/8