What are the solutions to the equation (3x-1)(x+3)=x(1+6x) ?

The solutions to the equation (3x-1)(x+3)=x(1+6x) are x=(7-sqrt(13))/6 ,x=(7+sqrt(13))/6

Find the zeros of (3x-1)(x+3)=x(1+6x)

The zeros of (3x-1)(x+3)=x(1+6x) are x=(7-sqrt(13))/6 ,x=(7+sqrt(13))/6

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(3 x - 1) (x + 3) = x (1 + 6 x)

x = \frac{7 - 13}{6}, x = \frac{7 + 13}{6}

Solution steps

(3 x - 1) (x + 3) = x (1 + 6 x)

Expand

(3 x - 1) (x + 3) : 3 x^{2} + 8 x - 3

Expand

x (1 + 6 x) : x + 6 x^{2}

3 x^{2} + 8 x - 3 = x + 6 x^{2}

Move

6 x^{2}

to the left side

- 3 x^{2} + 8 x - 3 = x

Move

x

to the left side

- 3 x^{2} + 7 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 3 ) ( - 3 )}{2 ( - 3 )}

7^{2} - 4 (- 3) (- 3) = 13

x_{1, 2} = \frac{- 7 \pm 13}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 7 + 13}{2 ( - 3 )}, x_{2} = \frac{- 7 - 13}{2 ( - 3 )}

x = \frac{- 7 + 13}{2 ( - 3 )} : \frac{7 - 13}{6}

x = \frac{- 7 - 13}{2 ( - 3 )} : \frac{7 + 13}{6}

The solutions to the quadratic equation are:

x = \frac{7 - 13}{6}, x = \frac{7 + 13}{6}

∣ 2 x - 5 ∣ \leq 7

5 ∣ 3 x + 9 ∣ + 10 = 35

2 x - 1 \geq 3

x^{2} + x - 2 = 0

25 n^{2} - 9

What are the solutions to the equation (3x-1)(x+3)=x(1+6x) ?
The solutions to the equation (3x-1)(x+3)=x(1+6x) are x=(7-sqrt(13))/6 ,x=(7+sqrt(13))/6
Find the zeros of (3x-1)(x+3)=x(1+6x)
The zeros of (3x-1)(x+3)=x(1+6x) are x=(7-sqrt(13))/6 ,x=(7+sqrt(13))/6