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

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

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

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

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

x = - \frac{2}{3}, x = 2

Decimal Notation

x = - 0.66666 \dots, x = 2

Solution steps

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

Expand

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

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

Move

4 x^{2}

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 (- 3) \cdot 4 = 8

x_{1, 2} = \frac{- 4 \pm 8}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 4 + 8}{2 ( - 3 )}, x_{2} = \frac{- 4 - 8}{2 ( - 3 )}

x = \frac{- 4 + 8}{2 ( - 3 )} : - \frac{2}{3}

x = \frac{- 4 - 8}{2 ( - 3 )} : 2

The solutions to the quadratic equation are:

x = - \frac{2}{3}, x = 2

4 x^{2} - 3 x - 7

x^{3} - 4 x^{2} - 19 x - 14

(6 n^{2} + n - 15)

- 4.9 t^{2} + 14.7 t = 0

x^{2} - 2 x = 4

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