Solve frac{sqrt{10}+3}{sqrt{10}-3}

$$\frac{ \sqrt{ 10 } +3 }{ \sqrt{ 10 } -3 }$$

Evaluate

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

Solution Steps

Rationalize the denominator of \frac{\sqrt{10}+3}{\sqrt{10}-3} by multiplying numerator and denominator by \sqrt{10}+3.

$$\frac{\left(\sqrt{10}+3\right)\left(\sqrt{10}+3\right)}{\left(\sqrt{10}-3\right)\left(\sqrt{10}+3\right)}$$

Consider \left(\sqrt{10}-3\right)\left(\sqrt{10}+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}.

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

Square \sqrt{10}. Square 3.

$$\frac{\left(\sqrt{10}+3\right)\left(\sqrt{10}+3\right)}{10-9}$$

Subtract 9 from 10 to get 1.

$$\frac{\left(\sqrt{10}+3\right)\left(\sqrt{10}+3\right)}{1}$$

Anything divided by one gives itself.

$$\left(\sqrt{10}+3\right)\left(\sqrt{10}+3\right)$$

Multiply \sqrt{10}+3 and \sqrt{10}+3 to get \left(\sqrt{10}+3\right)^{2}.

$$\left(\sqrt{10}+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{10}+3\right)^{2}.

$$\left(\sqrt{10}\right)^{2}+6\sqrt{10}+9$$

The square of \sqrt{10} is 10.

$$10+6\sqrt{10}+9$$

Add 10 and 9 to get 19.

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