Solve 1/3-sqrt{8}

$$\frac{ 1 }{ 3- \sqrt{ 8 } }$$

Evaluate

$$2\sqrt{2}+3\approx 5.828427125$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{1}{3-2\sqrt{2}}$$

Rationalize the denominator of \frac{1}{3-2\sqrt{2}} by multiplying numerator and denominator by 3+2\sqrt{2}.

$$\frac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}$$

Consider \left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

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

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

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

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

$$\frac{3+2\sqrt{2}}{9-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{3+2\sqrt{2}}{9-4\times 2}$$

Multiply 4 and 2 to get 8.

$$\frac{3+2\sqrt{2}}{9-8}$$

Subtract 8 from 9 to get 1.

$$\frac{3+2\sqrt{2}}{1}$$

Anything divided by one gives itself.

$$3+2\sqrt{2}$$