Question

Rationalize the denominator and simplify. StartFraction 5 minus StartRoot 3 EndRoot Over 4 plus 2 StartRoot 3 EndRoot EndFraction

Answer 1

Answer:

$\frac{13-7\sqrt{3}}{2}$

Step-by-step explanation:

We need to rationalize the denominator of $= \frac{5-\sqrt{3}}{4+2\sqrt{3}}$. For rationalizing we multiply the equation by $\frac{4-2\sqrt{3}}{4-2\sqrt{3}}$

So, solving

$= \frac{5-\sqrt{3}}{4+2\sqrt{3}}*\frac{4-2\sqrt{3}}{4-2\sqrt{3}} \\=\frac{(5-\sqrt{3})(4-2\sqrt{3})}{4+2\sqrt{3}*4-2\sqrt{3}}\\=\frac{(5-\sqrt{3})(4-2\sqrt{3})}{(4)^2-(2\sqrt{3})^2}\\= \frac{5(4-2\sqrt{3})-\sqrt{3}(4-2\sqrt{3})}{16-(4*3)}\\=\frac{20-10\sqrt{3}-4\sqrt{3}+2*3}{16-12}\\=\frac{20+6-14\sqrt{3}}{4}\\=\frac{26-14\sqrt{3}}{4}\\= \frac{2(13-7\sqrt{3})}{4}\\=\frac{13-7\sqrt{3}}{2}$