What are the solutions to the equation (1+x)^2-1=100x^2 ?

The solutions to the equation (1+x)^2-1=100x^2 are x=0,x= 2/99

Find the zeros of (1+x)^2-1=100x^2

The zeros of (1+x)^2-1=100x^2 are x=0,x= 2/99

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(1 + x)^{2} - 1 = 100 x^{2}

x = 0, x = \frac{2}{99}

Decimal Notation

x = 0, x = 0.02020 \dots

Solution steps

(1 + x)^{2} - 1 = 100 x^{2}

Expand

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

x^{2} + 2 x = 100 x^{2}

Move

100 x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 99 ) \cdot 0}{2 ( - 99 )}

2^{2} - 4 (- 99) \cdot 0 = 2

x_{1, 2} = \frac{- 2 \pm 2}{2 ( - 99 )}

Separate the solutions

x_{1} = \frac{- 2 + 2}{2 ( - 99 )}, x_{2} = \frac{- 2 - 2}{2 ( - 99 )}

x = \frac{- 2 + 2}{2 ( - 99 )} : 0

x = \frac{- 2 - 2}{2 ( - 99 )} : \frac{2}{99}

The solutions to the quadratic equation are:

x = 0, x = \frac{2}{99}

6 (x - 12) (x - 8) = 840

2 x^{2} - 19 x + 42 = 0

x^{2} - 3 (x + 4) = - 2 x

x^{2} + 5 x - 9 = 0

x^{2} - 18 x - 7 = 0

What are the solutions to the equation (1+x)^2-1=100x^2 ?
The solutions to the equation (1+x)^2-1=100x^2 are x=0,x= 2/99
Find the zeros of (1+x)^2-1=100x^2
The zeros of (1+x)^2-1=100x^2 are x=0,x= 2/99