What are the solutions to the equation 2t-t^3=-1 ?

The solutions to the equation 2t-t^3=-1 are t=-1,t=(1+sqrt(5))/2 ,t=(1-sqrt(5))/2

Find the zeros of 2t-t^3=-1

The zeros of 2t-t^3=-1 are t=-1,t=(1+sqrt(5))/2 ,t=(1-sqrt(5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 t - t^{3} = - 1

t = - 1, t = \frac{1 + 5}{2}, t = \frac{1 - 5}{2}

Decimal Notation

t = - 1, t = 1.61803 \dots, t = - 0.61803 \dots

Solution steps

2 t - t^{3} = - 1

Move

1

to the left side

2 t - t^{3} + 1 = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- t^{3} + 2 t + 1 = 0

Factor

- t^{3} + 2 t + 1 : - (t + 1) (t^{2} - t - 1)

- (t + 1) (t^{2} - t - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

t + 1 = 0 or t^{2} - t - 1 = 0

Solve

t + 1 = 0 : t = - 1

Solve

t^{2} - t - 1 = 0 : t = \frac{1 + 5}{2}, t = \frac{1 - 5}{2}

The solutions are

t = - 1, t = \frac{1 + 5}{2}, t = \frac{1 - 5}{2}

s^{3} + 6 s^{2} + 5 s + 10 = 0

x^{3} + 5 x^{2} + 9 x + 5 = 0

- x^{3} + 5 x^{2} - 8 x + 4

x^{3} + 3 x^{2} - 4 x - 12 = 0

x^{3} - \frac{5}{6} x^{2} + \frac{35}{6} x + 1 = 0

What are the solutions to the equation 2t-t^3=-1 ?
The solutions to the equation 2t-t^3=-1 are t=-1,t=(1+sqrt(5))/2 ,t=(1-sqrt(5))/2
Find the zeros of 2t-t^3=-1
The zeros of 2t-t^3=-1 are t=-1,t=(1+sqrt(5))/2 ,t=(1-sqrt(5))/2