Solve 8q^6-9q^3+1=0

$$8 q ^ { 6 } - 9 q ^ { 3 } + 1 = 0$$

Solve for q

$$q=\frac{1}{2}=0.5$$

$$q=1$$

Steps Using the Quadratic Formula

Substitute t for q^{3}.

$$8t^{2}-9t+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 8 for a, -9 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 8\times 1}}{2\times 8}$$

Do the calculations.

$$t=\frac{9±7}{16}$$

Solve the equation t=\frac{9±7}{16} when ± is plus and when ± is minus.

$$t=1 t=\frac{1}{8}$$

Since q=t^{3}, the solutions are obtained by evaluating q=\sqrt[3]{t} for each t.

$$q=1 q=\frac{1}{2}$$