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

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

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

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

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

x = 2, x = 0

Solution steps

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

Expand

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

Expand

(1 \cdot x + 1)^{2} : x^{2} + 2 x + 1

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

Move

1

to the left side

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

Move

2 x

to the left side

4 x^{2} - 6 x = x^{2}

Move

x^{2}

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

2 x^{2} - 7 x - 15 = 0

\frac{x ^{2}}{2} - \frac{3 x}{4} = 0

120000 = 350 x - 0.25 x^{2}

x^{2} + 6 x - 36 = 0

y^{2} + 4 y - 3 = 0

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