Solve 1/3-sqrt{8}-2sqrt{2}+6

$$\frac { 1 } { 3 - \sqrt { 8 } } - 2 \sqrt { 2 } + 6$$

Evaluate

$$9$$

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}.

$$\frac{1}{3-2\sqrt{2}}-2\sqrt{2}+6$$

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

$$\frac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}-2\sqrt{2}+6$$

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}.

$$\frac{3+2\sqrt{2}}{3^{2}-\left(-2\sqrt{2}\right)^{2}}-2\sqrt{2}+6$$

Calculate 3 to the power of 2 and get 9.

$$\frac{3+2\sqrt{2}}{9-\left(-2\sqrt{2}\right)^{2}}-2\sqrt{2}+6$$

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

$$\frac{3+2\sqrt{2}}{9-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}-2\sqrt{2}+6$$

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

$$\frac{3+2\sqrt{2}}{9-4\left(\sqrt{2}\right)^{2}}-2\sqrt{2}+6$$

The square of \sqrt{2} is 2.

$$\frac{3+2\sqrt{2}}{9-4\times 2}-2\sqrt{2}+6$$

Multiply 4 and 2 to get 8.

$$\frac{3+2\sqrt{2}}{9-8}-2\sqrt{2}+6$$

Subtract 8 from 9 to get 1.

$$\frac{3+2\sqrt{2}}{1}-2\sqrt{2}+6$$

Anything divided by one gives itself.

$$3+2\sqrt{2}-2\sqrt{2}+6$$

Combine 2\sqrt{2} and -2\sqrt{2} to get 0.

$$3+6$$

Add 3 and 6 to get 9.

$$9$$

Factor

$$3^{2}$$