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

The solutions to the equation 2x^2=x are x= 1/2 ,x=0

Find the zeros of 2x^2=x

The zeros of 2x^2=x are x= 1/2 ,x=0

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2 x^{2} = x

x = \frac{1}{2}, x = 0

Solution steps

2 x^{2} = x

Move

x

to the left side

2 x^{2} - x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = 0

(6 x^{\frac{1}{2}})^{2}

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

(- h^{4})^{5}

- 690

- 4 \leq 3 x - 5 < 12

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