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

The solutions to the equation (x+1)^2=4x^2 are x=-1/3 ,x=1

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

The zeros of (x+1)^2=4x^2 are x=-1/3 ,x=1

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

x = - \frac{1}{3}, x = 1

Decimal Notation

x = - 0.33333 \dots, x = 1

Solution steps

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

Expand

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

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

Move

4 x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 3 ) \cdot 1}{2 ( - 3 )}

2^{2} - 4 (- 3) \cdot 1 = 4

x_{1, 2} = \frac{- 2 \pm 4}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 2 + 4}{2 ( - 3 )}, x_{2} = \frac{- 2 - 4}{2 ( - 3 )}

x = \frac{- 2 + 4}{2 ( - 3 )} : - \frac{1}{3}

x = \frac{- 2 - 4}{2 ( - 3 )} : 1

The solutions to the quadratic equation are:

x = - \frac{1}{3}, x = 1

(n - 1)^{2} = 9

x^{2} + 3 x = 70

2 x + 3 = (x + 1)^{2}

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

\frac{2}{3} x^{2} - 6 = 0

What are the solutions to the equation (x+1)^2=4x^2 ?
The solutions to the equation (x+1)^2=4x^2 are x=-1/3 ,x=1
Find the zeros of (x+1)^2=4x^2
The zeros of (x+1)^2=4x^2 are x=-1/3 ,x=1