What are the solutions to the equation (x+1)(x-1)=15 ?

The solutions to the equation (x+1)(x-1)=15 are x=4,x=-4

Find the zeros of (x+1)(x-1)=15

The zeros of (x+1)(x-1)=15 are x=4,x=-4

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(x + 1) (x - 1) = 15

x = 4, x = - 4

Solution steps

(x + 1) (x - 1) = 15

Expand

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

x^{2} - 1 = 15

Move

15

to the left side

x^{2} - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 16 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 16) = 8

x_{1, 2} = \frac{- 0 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 8}{2 \cdot 1}, x_{2} = \frac{- 0 - 8}{2 \cdot 1}

x = \frac{- 0 + 8}{2 \cdot 1} : 4

x = \frac{- 0 - 8}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 4, x = - 4

3 x^{2} + 27 x + 60 = 0

x^{2} - 10 x - 16 = 0

(x - 2) (x + 6) = 16

16 - 2 x^{2} = 0

x^{2} - 10 x - 14 = 4

What are the solutions to the equation (x+1)(x-1)=15 ?
The solutions to the equation (x+1)(x-1)=15 are x=4,x=-4
Find the zeros of (x+1)(x-1)=15
The zeros of (x+1)(x-1)=15 are x=4,x=-4