Solve 3*frac{sqrt{37}-3}{6}*frac{sqrt{37}+3}{6}

$$3 \times \frac { \sqrt { 37 } - 3 } { 6 } \times \frac { \sqrt { 37 } + 3 } { 6 }$$

Evaluate

$$\frac{7}{3}\approx 2.333333333$$

Solution Steps

Cancel out 6, the greatest common factor in 3 and 6.

$$\frac{\sqrt{37}-3}{2}\times \frac{\sqrt{37}+3}{6}$$

Multiply \frac{\sqrt{37}-3}{2} times \frac{\sqrt{37}+3}{6} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(\sqrt{37}-3\right)\left(\sqrt{37}+3\right)}{2\times 6}$$

Consider \left(\sqrt{37}-3\right)\left(\sqrt{37}+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}.

$$\frac{\left(\sqrt{37}\right)^{2}-3^{2}}{2\times 6}$$

The square of \sqrt{37} is 37.

$$\frac{37-3^{2}}{2\times 6}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{37-9}{2\times 6}$$

Subtract 9 from 37 to get 28.

$$\frac{28}{2\times 6}$$

Multiply 2 and 6 to get 12.

$$\frac{28}{12}$$

Reduce the fraction \frac{28}{12} to lowest terms by extracting and canceling out 4.

$$\frac{7}{3}$$

Factor

$$\frac{7}{3} = 2\frac{1}{3} = 2.3333333333333335$$