Question

The square root of $\frac{{\left(3\frac{1}{4}\right)}^4-{\left(4\frac{1}{3}\right)}^4}{{\left(3\frac{1}{4}\right)}^2-{\left(4\frac{1}{3}\right)}^2}$ is

• A. $7\frac{1}{2}$
• B. $5\frac{5}{12}$
• C. $1\frac{1}{12}$
• D. $1\frac{7}{12}$

Answer 1

Answer: (B)

$5\frac{5}{12}$

$\frac{{\left(3\frac{1}{4}\right)}^4-{\left(4\frac{1}{3}\right)}^4}{{\left(3\frac{1}{4}\right)}^2-{\left(4\frac{1}{3}\right)}^2}$$=\frac{({\left(3\frac{1}{4}\right)}^2+{\left(4\frac{1}{3}\right)}^2)({\left(3\frac{1}{4}\right)}^2-{\left(4\frac{1}{3}\right)}^2)}{{\left(3\frac{1}{4}\right)}^2-{\left(4\frac{1}{3}\right)}^2}$ ....$[a^4-b^4=(a^2+b^2\left)\right(a^2-b^2\left)\right]$

$\Rightarrow {\left(\frac{13}{4}\right)}^2+{\left(\frac{13}{3}\right)}^2=\frac{169}{16}+\frac{169}{9}$

$\Rightarrow 169(\frac{1}{16}+\frac{1}{9})=169\times \left(\frac{9+16}{144}\right)$

$\Rightarrow \frac{169\times 25}{144}$

$\therefore \sqrt{\frac{169\times 25}{144}}$$=\frac{13\times 5}{12}=\frac{65}{12}=5\frac{5}{12}$