Question

Using the identity $a^2-b^2=(a+b)(a-b)$ solve ${\left(7\frac{3}{4}\right)}^2-{\left(2\frac{1}{4}\right)}^2$

Answer 1

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

We have $a^2-b^2=(a+b)(a-b)$

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

$=(\frac{31}{4}+\frac{9}{4})(\frac{31}{4}-\frac{9}{4})$

$=\frac{40}{4}\cdot \frac{22}{4}$

$=10\cdot \frac{11}{2}$

$=5\cdot 11$

${\left(7\frac{3}{4}\right)}^2-{\left(2\frac{1}{4}\right)}^2$$=55$