What are the solutions to the equation 9k^2-k-1=3k^2 ?

The solutions to the equation 9k^2-k-1=3k^2 are k= 1/2 ,k=-1/3

Find the zeros of 9k^2-k-1=3k^2

The zeros of 9k^2-k-1=3k^2 are k= 1/2 ,k=-1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9 k^{2} - k - 1 = 3 k^{2}

k = \frac{1}{2}, k = - \frac{1}{3}

Decimal Notation

k = 0.5, k = - 0.33333 \dots

Solution steps

9 k^{2} - k - 1 = 3 k^{2}

Move

3 k^{2}

to the left side

6 k^{2} - k - 1 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

(- 1)^{2} - 4 \cdot 6 (- 1) = 5

k_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 6}

Separate the solutions

k_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 6}, k_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 6}

k = \frac{- ( - 1 ) + 5}{2 \cdot 6} : \frac{1}{2}

k = \frac{- ( - 1 ) - 5}{2 \cdot 6} : - \frac{1}{3}

The solutions to the quadratic equation are:

k = \frac{1}{2}, k = - \frac{1}{3}

- \frac{4}{5} (7) - \frac{2}{5}

9 y - 5 = 2 y + 9

30 - 10 x > - 7 (x - 6)

10 \cdot 10

2 x + \frac{x}{3} \leq \frac{2}{5}

What are the solutions to the equation 9k^2-k-1=3k^2 ?
The solutions to the equation 9k^2-k-1=3k^2 are k= 1/2 ,k=-1/3
Find the zeros of 9k^2-k-1=3k^2
The zeros of 9k^2-k-1=3k^2 are k= 1/2 ,k=-1/3