Solve 21b^4-62b^2+25=0

$$21 b ^ { 4 } - 62 b ^ { 2 } + 25 = 0$$

Solve for b

$$b=-\frac{\sqrt{651-42\sqrt{109}}}{21}\approx -0.69417258$$

$$b=\frac{\sqrt{651-42\sqrt{109}}}{21}\approx 0.69417258$$

$$b = \frac{\sqrt{42 \sqrt{109} + 651}}{21} \approx 1.57178414$$

$$b = -\frac{\sqrt{42 \sqrt{109} + 651}}{21} \approx -1.57178414$$

Steps Using the Quadratic Formula

Substitute t for b^{2}.

$$21t^{2}-62t+25=0$$

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 21 for a, -62 for b, and 25 for c in the quadratic formula.

$$t=\frac{-\left(-62\right)±\sqrt{\left(-62\right)^{2}-4\times 21\times 25}}{2\times 21}$$

Do the calculations.

$$t=\frac{62±4\sqrt{109}}{42}$$

Solve the equation t=\frac{62±4\sqrt{109}}{42} when ± is plus and when ± is minus.

$$t=\frac{2\sqrt{109}+31}{21} t=\frac{31-2\sqrt{109}}{21}$$

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

$$b=\sqrt{\frac{2\sqrt{109}+31}{21}} b=-\sqrt{\frac{2\sqrt{109}+31}{21}} b=\sqrt{\frac{31-2\sqrt{109}}{21}} b=-\sqrt{\frac{31-2\sqrt{109}}{21}}$$