What are the solutions to the equation (3x-1)(4x+2)=(3x-1)(x+1) ?

The solutions to the equation (3x-1)(4x+2)=(3x-1)(x+1) are x= 1/3 ,x=-1/3

Find the zeros of (3x-1)(4x+2)=(3x-1)(x+1)

The zeros of (3x-1)(4x+2)=(3x-1)(x+1) are x= 1/3 ,x=-1/3

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(3 x - 1) (4 x + 2) = (3 x - 1) (x + 1)

x = \frac{1}{3}, x = - \frac{1}{3}

Decimal Notation

x = 0.33333 \dots, x = - 0.33333 \dots

Solution steps

(3 x - 1) (4 x + 2) = (3 x - 1) (x + 1)

Move

(3 x - 1) (x + 1)

to the left side

(3 x - 1) (4 x + 2) - (3 x - 1) (x + 1) = 0

Factor

(3 x - 1) (4 x + 2) - (3 x - 1) (x + 1) : (3 x - 1) (3 x + 1)

(3 x - 1) (3 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 1 = 0 or 3 x + 1 = 0

Solve

3 x - 1 = 0 : x = \frac{1}{3}

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3}, x = - \frac{1}{3}

- 2 x^{2} + 4 x = 5

p^{2} = 6 p + 27

7 x^{2} = 35 x

5 X^{2} - 20 = 0

9 m^{2} - 10 = 0

What are the solutions to the equation (3x-1)(4x+2)=(3x-1)(x+1) ?
The solutions to the equation (3x-1)(4x+2)=(3x-1)(x+1) are x= 1/3 ,x=-1/3
Find the zeros of (3x-1)(4x+2)=(3x-1)(x+1)
The zeros of (3x-1)(4x+2)=(3x-1)(x+1) are x= 1/3 ,x=-1/3