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

The solutions to the equation 9x^2-3x=1 are x=(1+sqrt(5))/6 ,x=(1-sqrt(5))/6

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

The zeros of 9x^2-3x=1 are x=(1+sqrt(5))/6 ,x=(1-sqrt(5))/6

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9 x^{2} - 3 x = 1

x = \frac{1 + 5}{6}, x = \frac{1 - 5}{6}

Decimal Notation

x = 0.53934 \dots, x = - 0.20601 \dots

Solution steps

9 x^{2} - 3 x = 1

Move

1

to the left side

9 x^{2} - 3 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 1 )}{2 \cdot 9}

(- 3)^{2} - 4 \cdot 9 (- 1) = 35

x_{1, 2} = \frac{- ( - 3 ) \pm 3 5}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 5}{2 \cdot 9}, x_{2} = \frac{- ( - 3 ) - 3 5}{2 \cdot 9}

x = \frac{- ( - 3 ) + 3 5}{2 \cdot 9} : \frac{1 + 5}{6}

x = \frac{- ( - 3 ) - 3 5}{2 \cdot 9} : \frac{1 - 5}{6}

The solutions to the quadratic equation are:

x = \frac{1 + 5}{6}, x = \frac{1 - 5}{6}

9 + 6 i

n - 10 = \frac{5}{6} n - 7 - \frac{1}{3} n

\frac{( x + 1 ) ( x - 1 )}{x} \leq 0

5 p - 3 \geq - 4 p - 7

\frac{4 x - 2}{3} - \frac{3 x - 1}{2} > 2

What are the solutions to the equation 9x^2-3x=1 ?
The solutions to the equation 9x^2-3x=1 are x=(1+sqrt(5))/6 ,x=(1-sqrt(5))/6
Find the zeros of 9x^2-3x=1
The zeros of 9x^2-3x=1 are x=(1+sqrt(5))/6 ,x=(1-sqrt(5))/6