Solve 1/7+3sqrt{2}*1/4

$$\frac { 1 } { 7 + 3 \sqrt { 2 } } \cdot \frac { 1 } { 4 }$$

Evaluate

$$\frac{7-3\sqrt{2}}{124}\approx 0.022236769$$

Solution Steps

Rationalize the denominator of \frac{1}{7+3\sqrt{2}} by multiplying numerator and denominator by 7-3\sqrt{2}.

$$\frac{7-3\sqrt{2}}{\left(7+3\sqrt{2}\right)\left(7-3\sqrt{2}\right)}\times \frac{1}{4}$$

Consider \left(7+3\sqrt{2}\right)\left(7-3\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{7-3\sqrt{2}}{7^{2}-\left(3\sqrt{2}\right)^{2}}\times \frac{1}{4}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{7-3\sqrt{2}}{49-\left(3\sqrt{2}\right)^{2}}\times \frac{1}{4}$$

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

$$\frac{7-3\sqrt{2}}{49-3^{2}\left(\sqrt{2}\right)^{2}}\times \frac{1}{4}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{7-3\sqrt{2}}{49-9\left(\sqrt{2}\right)^{2}}\times \frac{1}{4}$$

The square of \sqrt{2} is 2.

$$\frac{7-3\sqrt{2}}{49-9\times 2}\times \frac{1}{4}$$

Multiply 9 and 2 to get 18.

$$\frac{7-3\sqrt{2}}{49-18}\times \frac{1}{4}$$

Subtract 18 from 49 to get 31.

$$\frac{7-3\sqrt{2}}{31}\times \frac{1}{4}$$

Multiply \frac{7-3\sqrt{2}}{31} times \frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7-3\sqrt{2}}{31\times 4}$$

Multiply 31 and 4 to get 124.

$$\frac{7-3\sqrt{2}}{124}$$