Solve 2/4+3sqrt{2}+7/3-sqrt{2}-31/1+4sqrt{2}

$$\frac { 2 } { 4 + 3 \sqrt { 2 } } + \frac { 7 } { 3 - \sqrt { 2 } } - \frac { 31 } { 1 + 4 \sqrt { 2 } }$$

Evaluate

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Solution Steps

Rationalize the denominator of \frac{2}{4+3\sqrt{2}} by multiplying numerator and denominator by 4-3\sqrt{2}.

$$\frac{2\left(4-3\sqrt{2}\right)}{\left(4+3\sqrt{2}\right)\left(4-3\sqrt{2}\right)}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Consider \left(4+3\sqrt{2}\right)\left(4-3\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}.

$$\frac{2\left(4-3\sqrt{2}\right)}{4^{2}-\left(3\sqrt{2}\right)^{2}}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{2\left(4-3\sqrt{2}\right)}{16-\left(3\sqrt{2}\right)^{2}}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\frac{2\left(4-3\sqrt{2}\right)}{16-3^{2}\left(\sqrt{2}\right)^{2}}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{2\left(4-3\sqrt{2}\right)}{16-9\left(\sqrt{2}\right)^{2}}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

The square of \sqrt{2} is 2.

$$\frac{2\left(4-3\sqrt{2}\right)}{16-9\times 2}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{2\left(4-3\sqrt{2}\right)}{16-18}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Subtract 18 from 16 to get -2.

$$\frac{2\left(4-3\sqrt{2}\right)}{-2}+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Cancel out -2 and -2.

$$-\left(4-3\sqrt{2}\right)+\frac{7}{3-\sqrt{2}}-\frac{31}{1+4\sqrt{2}}$$

Rationalize the denominator of \frac{7}{3-\sqrt{2}} by multiplying numerator and denominator by 3+\sqrt{2}.

$$-\left(4-3\sqrt{2}\right)+\frac{7\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}-\frac{31}{1+4\sqrt{2}}$$

Consider \left(3-\sqrt{2}\right)\left(3+\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}.

$$-\left(4-3\sqrt{2}\right)+\frac{7\left(3+\sqrt{2}\right)}{3^{2}-\left(\sqrt{2}\right)^{2}}-\frac{31}{1+4\sqrt{2}}$$

Square 3. Square \sqrt{2}.

$$-\left(4-3\sqrt{2}\right)+\frac{7\left(3+\sqrt{2}\right)}{9-2}-\frac{31}{1+4\sqrt{2}}$$

Subtract 2 from 9 to get 7.

$$-\left(4-3\sqrt{2}\right)+\frac{7\left(3+\sqrt{2}\right)}{7}-\frac{31}{1+4\sqrt{2}}$$

Cancel out 7 and 7.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31}{1+4\sqrt{2}}$$

Rationalize the denominator of \frac{31}{1+4\sqrt{2}} by multiplying numerator and denominator by 1-4\sqrt{2}.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{\left(1+4\sqrt{2}\right)\left(1-4\sqrt{2}\right)}$$

Consider \left(1+4\sqrt{2}\right)\left(1-4\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}.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{1^{2}-\left(4\sqrt{2}\right)^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{1-\left(4\sqrt{2}\right)^{2}}$$

Expand \left(4\sqrt{2}\right)^{2}.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{1-4^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{1-16\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{1-16\times 2}$$

Multiply 16 and 2 to get 32.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{1-32}$$

Subtract 32 from 1 to get -31.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\frac{31\left(1-4\sqrt{2}\right)}{-31}$$

Cancel out -31 and -31.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}-\left(-\left(1-4\sqrt{2}\right)\right)$$

The opposite of -\left(1-4\sqrt{2}\right) is 1-4\sqrt{2}.

$$-\left(4-3\sqrt{2}\right)+3+\sqrt{2}+1-4\sqrt{2}$$

Add 3 and 1 to get 4.

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

Combine \sqrt{2} and -4\sqrt{2} to get -3\sqrt{2}.

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

To find the opposite of 4-3\sqrt{2}, find the opposite of each term.

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

The opposite of -3\sqrt{2} is 3\sqrt{2}.

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

Add -4 and 4 to get 0.

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

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

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Factor

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