Solve 1/b^2+1+9/4b^2=1

$$\frac { 1 } { b ^ { 2 } + 1 } + \frac { 9 } { 4 b ^ { 2 } } = 1$$

Solve for b

$$b=\sqrt{3}\approx 1.732050808$$

$$b=-\sqrt{3}\approx -1.732050808$$

Steps Using the Quadratic Formula

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4b^{2}\left(b^{2}+1\right), the least common multiple of b^{2}+1,4b^{2}.

$$4b^{2}+\left(b^{2}+1\right)\times 9=4b^{2}\left(b^{2}+1\right)$$

Use the distributive property to multiply b^{2}+1 by 9.

$$4b^{2}+9b^{2}+9=4b^{2}\left(b^{2}+1\right)$$

Combine 4b^{2} and 9b^{2} to get 13b^{2}.

$$13b^{2}+9=4b^{2}\left(b^{2}+1\right)$$

Use the distributive property to multiply 4b^{2} by b^{2}+1.

$$13b^{2}+9=4b^{4}+4b^{2}$$

Subtract 4b^{4} from both sides.

$$13b^{2}+9-4b^{4}=4b^{2}$$

Subtract 4b^{2} from both sides.

$$13b^{2}+9-4b^{4}-4b^{2}=0$$

Combine 13b^{2} and -4b^{2} to get 9b^{2}.

$$9b^{2}+9-4b^{4}=0$$

Substitute t for b^{2}.

$$-4t^{2}+9t+9=0$$

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 -4 for a, 9 for b, and 9 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-9±15}{-8}$$

Solve the equation t=\frac{-9±15}{-8} when ± is plus and when ± is minus.

$$t=-\frac{3}{4} t=3$$

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

$$b=\sqrt{3} b=-\sqrt{3}$$