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

The solutions to the equation 4x^2=-2x+1 are x=(-1+sqrt(5))/4 ,x=-(1+sqrt(5))/4

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

The zeros of 4x^2=-2x+1 are x=(-1+sqrt(5))/4 ,x=-(1+sqrt(5))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{- 1 + 5}{4}, x = - \frac{1 + 5}{4}

Decimal Notation

x = 0.30901 \dots, x = - 0.80901 \dots

Solution steps

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

Move

1

to the left side

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

Move

2 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 4 (- 1) = 25

x_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 2 + 2 5}{2 \cdot 4}, x_{2} = \frac{- 2 - 2 5}{2 \cdot 4}

x = \frac{- 2 + 2 5}{2 \cdot 4} : \frac{- 1 + 5}{4}

x = \frac{- 2 - 2 5}{2 \cdot 4} : - \frac{1 + 5}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 5}{4}, x = - \frac{1 + 5}{4}

x^{2} + 16 x + 3 = 0

y^{2} + 11 = 2

3 x^{2} - 3 x + 9 = 0

x^{2} + 16 x + 1 = 0

64 y^{2} + 7 = 32

What are the solutions to the equation 4x^2=-2x+1 ?
The solutions to the equation 4x^2=-2x+1 are x=(-1+sqrt(5))/4 ,x=-(1+sqrt(5))/4
Find the zeros of 4x^2=-2x+1
The zeros of 4x^2=-2x+1 are x=(-1+sqrt(5))/4 ,x=-(1+sqrt(5))/4