Solve frac{sqrt{2}}{2}=frac{sqrt{b^2-9}}{b}

$$\frac { \sqrt { 2 } } { 2 } = \frac { \sqrt { b ^ { 2 } - 9 } } { b }$$

Solve for b

$$b=3\sqrt{2}\approx 4.242640687$$

Solution Steps

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2b, the least common multiple of 2,b.

$$b\sqrt{2}=2\sqrt{b^{2}-9}$$

Subtract 2\sqrt{b^{2}-9} from both sides.

$$b\sqrt{2}-2\sqrt{b^{2}-9}=0$$

Subtract b\sqrt{2} from both sides of the equation.

$$-2\sqrt{b^{2}-9}=-b\sqrt{2}$$

Reorder the terms.

$$-2\sqrt{b^{2}-9}=-\sqrt{2}b$$

Square both sides of the equation.

$$\left(-2\sqrt{b^{2}-9}\right)^{2}=\left(-\sqrt{2}b\right)^{2}$$

Expand \left(-2\sqrt{b^{2}-9}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{b^{2}-9}\right)^{2}=\left(-\sqrt{2}b\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4\left(\sqrt{b^{2}-9}\right)^{2}=\left(-\sqrt{2}b\right)^{2}$$

Calculate \sqrt{b^{2}-9} to the power of 2 and get b^{2}-9.

$$4\left(b^{2}-9\right)=\left(-\sqrt{2}b\right)^{2}$$

Use the distributive property to multiply 4 by b^{2}-9.

$$4b^{2}-36=\left(-\sqrt{2}b\right)^{2}$$

Expand \left(-\sqrt{2}b\right)^{2}.

$$4b^{2}-36=\left(-1\right)^{2}\left(\sqrt{2}\right)^{2}b^{2}$$

Calculate -1 to the power of 2 and get 1.

$$4b^{2}-36=1\left(\sqrt{2}\right)^{2}b^{2}$$

The square of \sqrt{2} is 2.

$$4b^{2}-36=1\times 2b^{2}$$

Multiply 1 and 2 to get 2.

$$4b^{2}-36=2b^{2}$$

Subtract 2b^{2} from both sides.

$$4b^{2}-36-2b^{2}=0$$

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

$$2b^{2}-36=0$$

Add 36 to both sides. Anything plus zero gives itself.

$$2b^{2}=36$$

Divide both sides by 2.

$$b^{2}=\frac{36}{2}$$

Divide 36 by 2 to get 18.

$$b^{2}=18$$

Take the square root of both sides of the equation.

$$b=3\sqrt{2} b=-3\sqrt{2}$$

Substitute 3\sqrt{2} for b in the equation \frac{\sqrt{2}}{2}=\frac{\sqrt{b^{2}-9}}{b}.

$$\frac{\sqrt{2}}{2}=\frac{\sqrt{\left(3\sqrt{2}\right)^{2}-9}}{3\sqrt{2}}$$

Simplify. The value b=3\sqrt{2} satisfies the equation.

$$\frac{1}{2}\times 2^{\frac{1}{2}}=\frac{1}{2}\times 2^{\frac{1}{2}}$$

Substitute -3\sqrt{2} for b in the equation \frac{\sqrt{2}}{2}=\frac{\sqrt{b^{2}-9}}{b}.

$$\frac{\sqrt{2}}{2}=\frac{\sqrt{\left(-3\sqrt{2}\right)^{2}-9}}{-3\sqrt{2}}$$

Simplify. The value b=-3\sqrt{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{2}\times 2^{\frac{1}{2}}=-\frac{1}{2}\times 2^{\frac{1}{2}}$$

Equation -2\sqrt{b^{2}-9}=-\sqrt{2}b has a unique solution.

$$b=3\sqrt{2}$$