Question

Rationalize the denominator of the fraction below. What is the new denominator?

Answer 1

let's use the conjugate of the denominator, and multiply top and bottom by it, rationalizing the denominator is just a way of saying "get rid of that pesky radical at the bottom".

$\bf \cfrac{5}{3+\sqrt{6}}\cdot \cfrac{3-\sqrt{6}}{3-\sqrt{6}}\implies \cfrac{5(3-\sqrt{6})}{\underset{\textit{difference of squares}}{(3+\sqrt{6})(3-\sqrt{6})}}\implies \cfrac{15-5\sqrt{6}}{3^2-(\sqrt{6})^2} \\\\\\ \cfrac{15-5\sqrt{6}}{9-6}\implies \cfrac{15-5\sqrt{6}}{\boxed{3}}$

Answer 2

The new denominator of the given fraction is 3. So, option D is correct.

What is the rationalization of the denominator?

In a fraction, if the denominator has an imaginary number or a radical or a surd then it is eliminated by a process called rationalization.

• To rationalize a complex number that has a real part and an imaginary part, multiply both the numerator and denominator with the complex conjugate of the denominator.
• A complex number is in the form of (a + jb). So, its conjugate becomes (a - jb).
• To rationalize a complex number that has a real part and a radical (surd or irrational), multiply both numerator and denominator with the conjugate of the denominator.
• A complex number if it is in the form of (a + √b), then its conjugate is (a - √b).

Calculation:

The given fraction is $\frac{5}{3+\sqrt{6} }$

Since the denominator has a real part 3 and a radical √6, it is multiplied by its conjugate (3 - √6)

Multiplying both numerator and denominator by (3 - √6)

⇒ $\frac{5}{3+\sqrt{6} }*\frac{3-\sqrt{6} }{3-\sqrt{6} }$

⇒ $\frac{5(3-\sqrt{6} )}{(3+\sqrt{6})(3-\sqrt{6}) }$

Since the denominator is in the form of (a + b) (a - b) we can write a² - b².

So, (3 + √6)(3 - √6) = 3² - (√6)² = 9 - 6 = 3

⇒ $\frac{15-5\sqrt{6} }{3}$

Therefore, the new denominator is 3.

Learn more about rationalization here:

https://brainly.com/question/14261303

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