What would be the value of $a$ and $b$ in following rational expression?

Question

If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?

Answer 1

$$(5 + 2\sqrt{3}) = (7 + \sqrt{3})(a - \sqrt{3b})$$
$$=7a-7\sqrt b\sqrt 3 +a\sqrt 3-3\sqrt b$$
compare the coefficients of $\sqrt 3$ of both sides
we get$$7a-3\sqrt b=5$$ and $$a-7\sqrt b=2$$
solve for a and b
$a=29/46$ and $\sqrt b=-9/46$

Answer 2

$$\frac{5+2\sqrt{3}}{7+\sqrt{3}} = a - \sqrt{3b}$$ $$5+2\sqrt{3}=(a-\sqrt{3b})(7+\sqrt{3})$$ $(a-\sqrt{3b})(7+\sqrt{3})=7a+a\sqrt{3}-7\sqrt{3b}-3\sqrt{b}$
Now we have two equotation for two unknown $a$ and $b$:
$5 = 7a-3\sqrt{b}$ and
$2\sqrt{3}=(a-\sqrt{b})\sqrt{3} \implies 2=a-\sqrt{b}$.
Now
$\sqrt{b}=\frac{7a-5}{3}=a-2$
$7a-5=3a-6 \implies a = -\frac{1}{4}$
$\sqrt{b} = a - 2 = -\frac{1}{4} - \frac{8}{4}=-\frac{9}{4} \implies b = \frac{81}{16}$
So, $$a = -\frac{1}{4}, b = \frac{81}{16}$$

Answer 3

First rewrite $5+2\sqrt{3} = (a-\sqrt{3b})(7+\sqrt{3})$.

Now suppose $b=3n^2$, then we get $5+2\sqrt{3} = (a-3|n|)(7+\sqrt{3})$, which can not be solved for rational $a$ and $n$.

Next, suppose $b=n^2$, then we get $5+2\sqrt{3} = (a-|n|\sqrt{3})(7+\sqrt{3}) = 7a-3|n|+(a-7|n|)\sqrt{3}$ and hence $5=7a-3|n|$ and $2=a-7|n|$. This gives $9=-46|n|$, which has no solutions either for rational $a$ and $n$.

Other $b$ won't work either, since you'll be left with a $\sqrt{b}$. Thus your equation has no rational solutions.

Answer 4

What was written was $$\frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = a - \sqrt{3b},$$ but I'm going to boldly assume that's a typo and what was really intended was $$\frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = a - \sqrt{3}b.$$

Multiply the numerator and denominator by the conjugate of $7+\sqrt{3}$, which is $7-\sqrt{3}$: \begin{align} & \frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = \frac{(5 + 2\sqrt{3})(7 - \sqrt{3})}{(7 + \sqrt{3})(7 - \sqrt{3})} = \frac{29 + 9\sqrt{3}}{7^2-3} = \frac{29 + 9\sqrt{3}}{46} \\[10pt] & = \frac{29}{46} + \frac{9}{46}\sqrt{3} = \frac{29}{46} - \frac{-9}{46}\sqrt{3}. \end{align} So $a=\dfrac{29}{46}$ and $b=\dfrac{-9}{46}$.