What are the solutions to the equation (4x-1)^2+8x=4x+1 ?

The solutions to the equation (4x-1)^2+8x=4x+1 are x= 1/4 ,x=0

Find the zeros of (4x-1)^2+8x=4x+1

The zeros of (4x-1)^2+8x=4x+1 are x= 1/4 ,x=0

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(4 x - 1)^{2} + 8 x = 4 x + 1

x = \frac{1}{4}, x = 0

Solution steps

(4 x - 1)^{2} + 8 x = 4 x + 1

Expand

(4 x - 1)^{2} + 8 x : 16 x^{2} + 1

16 x^{2} + 1 = 4 x + 1

Move

1

to the left side

16 x^{2} = 4 x

Move

4 x

to the left side

16 x^{2} - 4 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 16 \cdot 0}{2 \cdot 16}

(- 4)^{2} - 4 \cdot 16 \cdot 0 = 4

x_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4}{2 \cdot 16}, x_{2} = \frac{- ( - 4 ) - 4}{2 \cdot 16}

x = \frac{- ( - 4 ) + 4}{2 \cdot 16} : \frac{1}{4}

x = \frac{- ( - 4 ) - 4}{2 \cdot 16} : 0

The solutions to the quadratic equation are:

x = \frac{1}{4}, x = 0

- 20 - 3 x + 3 x^{2} = 0

(2 x - 5)^{2} - 81 = 0

4 x^{2} + 23 x - 72 = 0

x + 30 = 4 x^{2} + 20 x + 25

16 - 24 x + 9 x^{2} = 0

What are the solutions to the equation (4x-1)^2+8x=4x+1 ?
The solutions to the equation (4x-1)^2+8x=4x+1 are x= 1/4 ,x=0
Find the zeros of (4x-1)^2+8x=4x+1
The zeros of (4x-1)^2+8x=4x+1 are x= 1/4 ,x=0