Question

Rationalize the denominator 3-sqrt(6)÷4+sqrt(5)

Answer 1

Answer:

  $\dfrac{12-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{11}$

Step-by-step explanation:

You have written 3 - ((√6)/4) + √5. The only denominator is 4, which is already rational. If you were to use appropriate grouping symbols (parentheses) we might expect to see ...

  (3 -√6)/(4 +√5)

The denominator is rationalized by taking advantage of the factoring of the difference of squares:

  a² - b² = (a -b)(a +b)

The denominator will be rational if we can square the value √5. By choosing a multiplier of (4-√5), we can do that:

$\dfrac{3-\sqrt{6}}{4+\sqrt{5}}=\dfrac{(3-\sqrt{6})(4-\sqrt{5})}{(4+\sqrt{5})(4-\sqrt{5})}\\\\=\dfrac{3\cdot 4-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{4^2-(\sqrt{5})^2}=\dfrac{12-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{11}$