Solve k^2-1/8=QRightarrowk^2-1=8Q

$$\frac { k ^ { 2 } - 1 } { 8 } = Q \Rightarrow k ^ { 2 } - 1 = 8 Q$$

Solve for Q

$$Q=\frac{k^{2}-1}{8}$$

Solution Steps

Divide each term of k^{2}-1 by 8 to get \frac{1}{8}k^{2}-\frac{1}{8}.

$$\frac{1}{8}k^{2}-\frac{1}{8}=Q$$

Swap sides so that all variable terms are on the left hand side.

$$Q=\frac{1}{8}k^{2}-\frac{1}{8}$$

Solve for k

$$k=\sqrt{8Q+1}$$

$$k=-\sqrt{8Q+1}\text{, }Q\geq -\frac{1}{8}$$