Solve frac{4-sqrt{8}}{2+sqrt{8}}

$$\frac{ 4- \sqrt{ 8 } }{ 2+ \sqrt{ 8 } }$$

Evaluate

$$3\sqrt{2}-4\approx 0.242640687$$

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}.

$$\frac{4-2\sqrt{2}}{2+\sqrt{8}}$$

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}.

$$\frac{4-2\sqrt{2}}{2+2\sqrt{2}}$$

Rationalize the denominator of \frac{4-2\sqrt{2}}{2+2\sqrt{2}} by multiplying numerator and denominator by 2-2\sqrt{2}.

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

Consider \left(2+2\sqrt{2}\right)\left(2-2\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}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

$$\frac{\left(4-2\sqrt{2}\right)\left(2-2\sqrt{2}\right)}{4-4\times 2}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(4-2\sqrt{2}\right)\left(2-2\sqrt{2}\right)}{4-8}$$

Subtract 8 from 4 to get -4.

$$\frac{\left(4-2\sqrt{2}\right)\left(2-2\sqrt{2}\right)}{-4}$$

Apply the distributive property by multiplying each term of 4-2\sqrt{2} by each term of 2-2\sqrt{2}.

$$\frac{8-8\sqrt{2}-4\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{-4}$$

Combine -8\sqrt{2} and -4\sqrt{2} to get -12\sqrt{2}.

$$\frac{8-12\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{-4}$$

The square of \sqrt{2} is 2.

$$\frac{8-12\sqrt{2}+4\times 2}{-4}$$

Multiply 4 and 2 to get 8.

$$\frac{8-12\sqrt{2}+8}{-4}$$

Add 8 and 8 to get 16.

$$\frac{16-12\sqrt{2}}{-4}$$

Divide each term of 16-12\sqrt{2} by -4 to get -4+3\sqrt{2}.

$$-4+3\sqrt{2}$$