What are the solutions to the equation (x+1)^3-(x-1)^3=7x ?

The solutions to the equation (x+1)^3-(x-1)^3=7x are x= 2/3 ,x= 1/2

Find the zeros of (x+1)^3-(x-1)^3=7x

The zeros of (x+1)^3-(x-1)^3=7x are x= 2/3 ,x= 1/2

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(x + 1)^{3} - (x - 1)^{3} = 7 x

x = \frac{2}{3}, x = \frac{1}{2}

Solution steps

(x + 1)^{3} - (x - 1)^{3} = 7 x

Expand

(x + 1)^{3} - (x - 1)^{3} : 6 x^{2} + 2

6 x^{2} + 2 = 7 x

Move

7 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 2}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 \cdot 2 = 1

x_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 6}, x_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 6}

x = \frac{- ( - 7 ) + 1}{2 \cdot 6} : \frac{2}{3}

x = \frac{- ( - 7 ) - 1}{2 \cdot 6} : \frac{1}{2}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = \frac{1}{2}

1^{2} + 3^{2} = x^{2}

4 x^{2} - 15 x = 0

12 x^{2} - 112 x + 180 = 0

\frac{x ^{2}}{8} = \frac{1}{2}

x^{2} + 15 x - 2 = 0

What are the solutions to the equation (x+1)^3-(x-1)^3=7x ?
The solutions to the equation (x+1)^3-(x-1)^3=7x are x= 2/3 ,x= 1/2
Find the zeros of (x+1)^3-(x-1)^3=7x
The zeros of (x+1)^3-(x-1)^3=7x are x= 2/3 ,x= 1/2