Question
Find rational numbers a and b such that:$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=\text{a}+\text{b}\sqrt{3}$
✓
Answer
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=\text{a}+\text{b}\sqrt{3}$we have,
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$
$=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times\frac{7-4\sqrt{3}}{7-4\sqrt{3}}$
$=\frac{5\times7-5\times4\sqrt{3}+2\sqrt{3}\times7-2\sqrt{3}\times4\sqrt{3}}{(7)^2-\big(4\sqrt{3}\big)^2}$
$=\frac{35-20\sqrt{3}+14\sqrt{3}-8\times3}{49-16\times3}$
$=\frac{35-6\sqrt{3}-24}{49-48}$
$=\frac{11-6\sqrt{3}}{1}$
$=11-6\sqrt{3}$
Now,
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=\text{a}+\text{b}\sqrt{3}$
$\Rightarrow\text{a}=11$ and $\text{b}=-6$
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$
$=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times\frac{7-4\sqrt{3}}{7-4\sqrt{3}}$
$=\frac{5\times7-5\times4\sqrt{3}+2\sqrt{3}\times7-2\sqrt{3}\times4\sqrt{3}}{(7)^2-\big(4\sqrt{3}\big)^2}$
$=\frac{35-20\sqrt{3}+14\sqrt{3}-8\times3}{49-16\times3}$
$=\frac{35-6\sqrt{3}-24}{49-48}$
$=\frac{11-6\sqrt{3}}{1}$
$=11-6\sqrt{3}$
Now,
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=\text{a}+\text{b}\sqrt{3}$
$\Rightarrow\text{a}=11$ and $\text{b}=-6$
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