Solve 9m^4-40m^2+11=0

$$9 m ^ { 4 } - 40 m ^ { 2 } + 11 = 0$$

Solve for m

$$m=-\frac{\sqrt{20-\sqrt{301}}}{3}\approx -0.542693737$$

$$m=\frac{\sqrt{20-\sqrt{301}}}{3}\approx 0.542693737$$

$$m = \frac{\sqrt{\sqrt{301} + 20}}{3} \approx 2.037137195$$

$$m = -\frac{\sqrt{\sqrt{301} + 20}}{3} \approx -2.037137195$$

Steps Using the Quadratic Formula

Substitute t for m^{2}.

$$9t^{2}-40t+11=0$$

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 9 for a, -40 for b, and 11 for c in the quadratic formula.

$$t=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 9\times 11}}{2\times 9}$$

Do the calculations.

$$t=\frac{40±2\sqrt{301}}{18}$$

Solve the equation t=\frac{40±2\sqrt{301}}{18} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{301}+20}{9} t=\frac{20-\sqrt{301}}{9}$$

Since m=t^{2}, the solutions are obtained by evaluating m=±\sqrt{t} for each t.

$$m=\frac{\sqrt{\sqrt{301}+20}}{3} m=-\frac{\sqrt{\sqrt{301}+20}}{3} m=\frac{\sqrt{20-\sqrt{301}}}{3} m=-\frac{\sqrt{20-\sqrt{301}}}{3}$$