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

The solutions to the equation 2(m-1)(1+m)=m are m=(1+sqrt(17))/4 ,m=(1-sqrt(17))/4

Find the zeros of 2(m-1)(1+m)=m

The zeros of 2(m-1)(1+m)=m are m=(1+sqrt(17))/4 ,m=(1-sqrt(17))/4

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2 (m - 1) (1 + m) = m

m = \frac{1 + 17}{4}, m = \frac{1 - 17}{4}

Decimal Notation

m = 1.28077 \dots, m = - 0.78077 \dots

Solution steps

2 (m - 1) (1 + m) = m

Expand

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

2 m^{2} - 2 = m

Move

m

to the left side

2 m^{2} - 2 - m = 0

Write in the standard form

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

2 m^{2} - m - 2 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 2) = 17

m_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 \cdot 2}

Separate the solutions

m_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 2}, m_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 2}

m = \frac{- ( - 1 ) + 17}{2 \cdot 2} : \frac{1 + 17}{4}

m = \frac{- ( - 1 ) - 17}{2 \cdot 2} : \frac{1 - 17}{4}

The solutions to the quadratic equation are:

m = \frac{1 + 17}{4}, m = \frac{1 - 17}{4}

x, x^{2} + 4 x + 13 = 0

16.4 = 8.1 t + 4.9 t^{2}

480 = (6 x - 4) (4 x - 4) (2)

x^{2} + (- 282) = 51

x (x + 1) - 2 (x + 1) = 0

What are the solutions to the equation 2(m-1)(1+m)=m ?
The solutions to the equation 2(m-1)(1+m)=m are m=(1+sqrt(17))/4 ,m=(1-sqrt(17))/4
Find the zeros of 2(m-1)(1+m)=m
The zeros of 2(m-1)(1+m)=m are m=(1+sqrt(17))/4 ,m=(1-sqrt(17))/4