Solve 1/6+sqrt{35}+1/6-sqrt{35}

$$\frac{ 1 }{ 6+ \sqrt{ 35 } } + \frac{ 1 }{ 6- \sqrt{ 35 } }$$

Evaluate

$$12$$

Solution Steps

Rationalize the denominator of \frac{1}{6+\sqrt{35}} by multiplying numerator and denominator by 6-\sqrt{35}.

$$\frac{6-\sqrt{35}}{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}+\frac{1}{6-\sqrt{35}}$$

Consider \left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{6-\sqrt{35}}{6^{2}-\left(\sqrt{35}\right)^{2}}+\frac{1}{6-\sqrt{35}}$$

Square 6. Square \sqrt{35}.

$$\frac{6-\sqrt{35}}{36-35}+\frac{1}{6-\sqrt{35}}$$

Subtract 35 from 36 to get 1.

$$\frac{6-\sqrt{35}}{1}+\frac{1}{6-\sqrt{35}}$$

Anything divided by one gives itself.

$$6-\sqrt{35}+\frac{1}{6-\sqrt{35}}$$

Rationalize the denominator of \frac{1}{6-\sqrt{35}} by multiplying numerator and denominator by 6+\sqrt{35}.

$$6-\sqrt{35}+\frac{6+\sqrt{35}}{\left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right)}$$

Consider \left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$6-\sqrt{35}+\frac{6+\sqrt{35}}{6^{2}-\left(\sqrt{35}\right)^{2}}$$

Square 6. Square \sqrt{35}.

$$6-\sqrt{35}+\frac{6+\sqrt{35}}{36-35}$$

Subtract 35 from 36 to get 1.

$$6-\sqrt{35}+\frac{6+\sqrt{35}}{1}$$

Anything divided by one gives itself.

$$6-\sqrt{35}+6+\sqrt{35}$$

Add 6 and 6 to get 12.

$$12-\sqrt{35}+\sqrt{35}$$

Combine -\sqrt{35} and \sqrt{35} to get 0.

$$12$$

Factor

$$2^{2}\times 3$$