What are the solutions to the equation (k+1)(k-1)=8 ?

The solutions to the equation (k+1)(k-1)=8 are k=3,k=-3

Find the zeros of (k+1)(k-1)=8

The zeros of (k+1)(k-1)=8 are k=3,k=-3

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(k + 1) (k - 1) = 8

k = 3, k = - 3

Solution steps

(k + 1) (k - 1) = 8

Expand

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

k^{2} - 1 = 8

Move

8

to the left side

k^{2} - 9 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 9) = 6

k_{1, 2} = \frac{- 0 \pm 6}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 0 + 6}{2 \cdot 1}, k_{2} = \frac{- 0 - 6}{2 \cdot 1}

k = \frac{- 0 + 6}{2 \cdot 1} : 3

k = \frac{- 0 - 6}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

k = 3, k = - 3

x^{2} + \frac{2}{3} a x + (- \frac{a ^{2}}{2} + 1) = 0

x (x - 5) - 4 (3 - x) = 0

a, (a^{2} + 1) (b^{2} + 4) = 10 ab - 5

1 = \frac{x ^{2}}{4}

x^{2} + 20 x + 28 = 9

What are the solutions to the equation (k+1)(k-1)=8 ?
The solutions to the equation (k+1)(k-1)=8 are k=3,k=-3
Find the zeros of (k+1)(k-1)=8
The zeros of (k+1)(k-1)=8 are k=3,k=-3