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

The solutions to the equation (3x+1)^2-(3x-1)^2=x^2 are x=12,x=0

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

The zeros of (3x+1)^2-(3x-1)^2=x^2 are x=12,x=0

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(3 x + 1)^{2} - (3 x - 1)^{2} = x^{2}

x = 12, x = 0

Solution steps

(3 x + 1)^{2} - (3 x - 1)^{2} = x^{2}

Expand

(3 x + 1)^{2} - (3 x - 1)^{2} : 12 x

12 x = x^{2}

Switch sides

x^{2} = 12 x

Move

12 x

to the left side

x^{2} - 12 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 0 = 12

x_{1, 2} = \frac{- ( - 12 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 12}{2 \cdot 1}

x = \frac{- ( - 12 ) + 12}{2 \cdot 1} : 12

x = \frac{- ( - 12 ) - 12}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 12, x = 0

2 x^{2} + 3.7 x + 0.305 = 0

x^{2} - 54 = 3 x

- 3 x^{2} - 30 x + 117 = 0

\frac{9 x ^{2} - 7 x}{4} = 0

\frac{3}{4} x^{2} = 0

What are the solutions to the equation (3x+1)^2-(3x-1)^2=x^2 ?
The solutions to the equation (3x+1)^2-(3x-1)^2=x^2 are x=12,x=0
Find the zeros of (3x+1)^2-(3x-1)^2=x^2
The zeros of (3x+1)^2-(3x-1)^2=x^2 are x=12,x=0