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

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

Find the zeros of 2x^2+2x=4x-1

The zeros of 2x^2+2x=4x-1 are x= 1/2+i 1/2 ,x= 1/2-i 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{1}{2} + i \frac{1}{2}, x = \frac{1}{2} - i \frac{1}{2}

Solution steps

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

Move

1

to the left side

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

Move

4 x

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 1 : 2 i

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 2 i}{2 \cdot 2} : \frac{1}{2} + i \frac{1}{2}

x = \frac{- ( - 2 ) - 2 i}{2 \cdot 2} : \frac{1}{2} - i \frac{1}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{1}{2}, x = \frac{1}{2} - i \frac{1}{2}

\frac{2}{3} x + 7 = \frac{5}{6}

∣ 4 v + 7 ∣ = ∣ 4 v + 5 ∣

6 s t^{2} - 18 s t + 6 t^{4} + 18 t^{3}

lo g_{4} (\frac{1}{2}) = x

lo g_{3} (- 27)

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