Solve frac{3sqrt{2}+4}{3sqrt{2}-4}

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

Evaluate

$$12\sqrt{2}+17\approx 33.970562748$$

Solution Steps

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

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

Consider \left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Multiply 3\sqrt{2}+4 and 3\sqrt{2}+4 to get \left(3\sqrt{2}+4\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3\sqrt{2}+4\right)^{2}.

$$\frac{9\left(\sqrt{2}\right)^{2}+24\sqrt{2}+16}{\left(3\sqrt{2}\right)^{2}-4^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{9\times 2+24\sqrt{2}+16}{\left(3\sqrt{2}\right)^{2}-4^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{18+24\sqrt{2}+16}{\left(3\sqrt{2}\right)^{2}-4^{2}}$$

Add 18 and 16 to get 34.

$$\frac{34+24\sqrt{2}}{\left(3\sqrt{2}\right)^{2}-4^{2}}$$

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

$$\frac{34+24\sqrt{2}}{3^{2}\left(\sqrt{2}\right)^{2}-4^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{34+24\sqrt{2}}{9\left(\sqrt{2}\right)^{2}-4^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{34+24\sqrt{2}}{9\times 2-4^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{34+24\sqrt{2}}{18-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{34+24\sqrt{2}}{18-16}$$

Subtract 16 from 18 to get 2.

$$\frac{34+24\sqrt{2}}{2}$$

Divide each term of 34+24\sqrt{2} by 2 to get 17+12\sqrt{2}.

$$17+12\sqrt{2}$$