Solve frac{1+3sqrt{2}}{sqrt{2}-1}

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

Evaluate

$$4\sqrt{2}+7\approx 12.656854249$$

Solution Steps

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

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

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}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

Apply the distributive property by multiplying each term of 1+3\sqrt{2} by each term of \sqrt{2}+1.

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

Add 1 and 6 to get 7.

$$\sqrt{2}+7+3\sqrt{2}$$

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

$$4\sqrt{2}+7$$