Question

If $\frac{{\left(3\frac{1}{4}\right)}^4-{\left(4\frac{1}{3}\right)}^4}{{\left(3\frac{1}{4}\right)}^2-{\left(4\frac{1}{3}\right)}^2}={\left(\frac{13a}{12}\right)}^2$,find $a$.

Answer 1

$\frac{{\left(3\frac{1}{4}\right)}^4-{\left(4\frac{1}{3}\right)}^4}{{\left(3\frac{1}{4}\right)}^2-{\left(4\frac{1}{3}\right)}^2}={\left(\frac{13a}{12}\right)}^2$

$\Rightarrow \frac{\frac{28561}{256}-\frac{28561}{81}}{\frac{169}{16}-\frac{169}{9}}=\frac{169a^2}{144}$

$\Rightarrow \frac{111.56-352.60}{10.56-18.77}=\frac{169a2}{144}$

$\Rightarrow \frac{-241.04}{-8.21}=\frac{169a^2}{144}$

$\Rightarrow 29.35=\frac{169a^2}{144}$

$\Rightarrow 169a^2=29.35\times 144$

$\Rightarrow 169a^2=4226.6$

$\Rightarrow a^2=4226.6÷169$

$\Rightarrow a^2=25$

$\Rightarrow a=5$