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

The solutions to the equation (x+1)^2=4x are x=1

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

The zeros of (x+1)^2=4x are x=1

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

x = 1

Solution steps

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

Expand

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

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

Move

4 x

to the left side

x^{2} - 2 x + 1 = 0

Solve with the quadratic formula

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

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

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

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

- 6 x^{2} + 5 x - 1 = 0

64 r^{2} + 1 = 26

(3 x - 2)^{2} = 7

5 x^{2} + 6 = 6

(4 x + 7)^{2} = (7 x + 4)^{2}

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