What are the solutions to the equation (1-2x)(1+2x)=3x ?

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

Find the zeros of (1-2x)(1+2x)=3x

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

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(1 - 2 x) (1 + 2 x) = 3 x

x = - 1, x = \frac{1}{4}

Decimal Notation

x = - 1, x = 0.25

Solution steps

(1 - 2 x) (1 + 2 x) = 3 x

Expand

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

1 - 4 x^{2} = 3 x

Move

3 x

to the left side

1 - 4 x^{2} - 3 x = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 4) \cdot 1 = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 ( - 4 )}, x_{2} = \frac{- ( - 3 ) - 5}{2 ( - 4 )}

x = \frac{- ( - 3 ) + 5}{2 ( - 4 )} : - 1

x = \frac{- ( - 3 ) - 5}{2 ( - 4 )} : \frac{1}{4}

The solutions to the quadratic equation are:

x = - 1, x = \frac{1}{4}

3 x^{2} - 30 = 30

\frac{3 x}{8} - \frac{x ^{2}}{8} = 0

x^{2} + 36 x + 332 = 0

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

13 x^{2} - 27 = 0

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