Solve frac{sqrt{30}+3sqrt{3}}{sqrt{30}-3sqrt{3}}

$$\frac { \sqrt { 30 } + 3 \sqrt { 3 } } { \sqrt { 30 } - 3 \sqrt { 3 } }$$

Evaluate

$$6\sqrt{10}+19\approx 37.973665961$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{30}+3\sqrt{3}}{\sqrt{30}-3\sqrt{3}} by multiplying numerator and denominator by \sqrt{30}+3\sqrt{3}.

$$\frac{\left(\sqrt{30}+3\sqrt{3}\right)\left(\sqrt{30}+3\sqrt{3}\right)}{\left(\sqrt{30}-3\sqrt{3}\right)\left(\sqrt{30}+3\sqrt{3}\right)}$$

Consider \left(\sqrt{30}-3\sqrt{3}\right)\left(\sqrt{30}+3\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{30}+3\sqrt{3}\right)\left(\sqrt{30}+3\sqrt{3}\right)}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Multiply \sqrt{30}+3\sqrt{3} and \sqrt{30}+3\sqrt{3} to get \left(\sqrt{30}+3\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{30}+3\sqrt{3}\right)^{2}}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{30}+3\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{30}\right)^{2}+6\sqrt{30}\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

The square of \sqrt{30} is 30.

$$\frac{30+6\sqrt{30}\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Factor 30=3\times 10. Rewrite the square root of the product \sqrt{3\times 10} as the product of square roots \sqrt{3}\sqrt{10}.

$$\frac{30+6\sqrt{3}\sqrt{10}\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{30+6\times 3\sqrt{10}+9\left(\sqrt{3}\right)^{2}}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Multiply 6 and 3 to get 18.

$$\frac{30+18\sqrt{10}+9\left(\sqrt{3}\right)^{2}}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{30+18\sqrt{10}+9\times 3}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Multiply 9 and 3 to get 27.

$$\frac{30+18\sqrt{10}+27}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

Add 30 and 27 to get 57.

$$\frac{57+18\sqrt{10}}{\left(\sqrt{30}\right)^{2}-\left(-3\sqrt{3}\right)^{2}}$$

The square of \sqrt{30} is 30.

$$\frac{57+18\sqrt{10}}{30-\left(-3\sqrt{3}\right)^{2}}$$

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

$$\frac{57+18\sqrt{10}}{30-\left(-3\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$\frac{57+18\sqrt{10}}{30-9\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{57+18\sqrt{10}}{30-9\times 3}$$

Multiply 9 and 3 to get 27.

$$\frac{57+18\sqrt{10}}{30-27}$$

Subtract 27 from 30 to get 3.

$$\frac{57+18\sqrt{10}}{3}$$

Divide each term of 57+18\sqrt{10} by 3 to get 19+6\sqrt{10}.

$$19+6\sqrt{10}$$