Solve 1/sqrt{2+1}-7/sqrt{8-1}+sqrt{8}

$$\frac { 1 } { \sqrt { 2 } + 1 } - \frac { 7 } { \sqrt { 8 } - 1 } + \sqrt { 8 }$$

Evaluate

$$\sqrt{2}-2\approx -0.585786438$$

Solution Steps

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

$$\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}-\frac{7}{\sqrt{8}-1}+\sqrt{8}$$

Consider \left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{2}-1}{\left(\sqrt{2}\right)^{2}-1^{2}}-\frac{7}{\sqrt{8}-1}+\sqrt{8}$$

Square \sqrt{2}. Square 1.

$$\frac{\sqrt{2}-1}{2-1}-\frac{7}{\sqrt{8}-1}+\sqrt{8}$$

Subtract 1 from 2 to get 1.

$$\frac{\sqrt{2}-1}{1}-\frac{7}{\sqrt{8}-1}+\sqrt{8}$$

Anything divided by one gives itself.

$$\sqrt{2}-1-\frac{7}{\sqrt{8}-1}+\sqrt{8}$$

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}.

$$\sqrt{2}-1-\frac{7}{2\sqrt{2}-1}+\sqrt{8}$$

Rationalize the denominator of \frac{7}{2\sqrt{2}-1} by multiplying numerator and denominator by 2\sqrt{2}+1.

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)}+\sqrt{8}$$

Consider \left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{\left(2\sqrt{2}\right)^{2}-1^{2}}+\sqrt{8}$$

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

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{2^{2}\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt{8}$$

Calculate 2 to the power of 2 and get 4.

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{4\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt{8}$$

The square of \sqrt{2} is 2.

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{4\times 2-1^{2}}+\sqrt{8}$$

Multiply 4 and 2 to get 8.

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{8-1^{2}}+\sqrt{8}$$

Calculate 1 to the power of 2 and get 1.

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{8-1}+\sqrt{8}$$

Subtract 1 from 8 to get 7.

$$\sqrt{2}-1-\frac{7\left(2\sqrt{2}+1\right)}{7}+\sqrt{8}$$

Cancel out 7 and 7.

$$\sqrt{2}-1-\left(2\sqrt{2}+1\right)+\sqrt{8}$$

To find the opposite of 2\sqrt{2}+1, find the opposite of each term.

$$\sqrt{2}-1-2\sqrt{2}-1+\sqrt{8}$$

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

$$-\sqrt{2}-1-1+\sqrt{8}$$

Subtract 1 from -1 to get -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}.

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

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

$$\sqrt{2}-2$$