#3**+16 **

Yes you have to rationalise the denominator.

remember $$(a-b)(a+b)=a^2-b^2$$ This is a difference of 2 squares.

oh, a+b and a-b are called a congugates of one another. That is because they are the same except for the +- sign in the middle.

When rationalizing denominators like this you use this fact. because it a or b is a surd then a^{2} and b^{2} are not!

$$\frac{1}{\sqrt2+\sqrt3}\\\\

=\frac{1}{\sqrt2+\sqrt3} \times \frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}\\\\

=\frac{(\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)} \\\\

=\frac{(\sqrt2-\sqrt3)}{(\sqrt2)^2-(\sqrt3)^2} \\\\

=\frac{(\sqrt2-\sqrt3)}{2-3} \\\\

=\frac{(\sqrt2-\sqrt3)}{-1} \\\\

=\frac{(\sqrt3-\sqrt2)}{+1} \\\\

=\sqrt3-\sqrt2\\$$

Melody
Aug 5, 2014

#1**+8 **

Does it involve division too?

Which one are you talking about? About Problem 21, or about D?

DragonSlayer554
Aug 5, 2014

#3**+16 **

Best Answer

Yes you have to rationalise the denominator.

remember $$(a-b)(a+b)=a^2-b^2$$ This is a difference of 2 squares.

oh, a+b and a-b are called a congugates of one another. That is because they are the same except for the +- sign in the middle.

When rationalizing denominators like this you use this fact. because it a or b is a surd then a^{2} and b^{2} are not!

$$\frac{1}{\sqrt2+\sqrt3}\\\\

=\frac{1}{\sqrt2+\sqrt3} \times \frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}\\\\

=\frac{(\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)} \\\\

=\frac{(\sqrt2-\sqrt3)}{(\sqrt2)^2-(\sqrt3)^2} \\\\

=\frac{(\sqrt2-\sqrt3)}{2-3} \\\\

=\frac{(\sqrt2-\sqrt3)}{-1} \\\\

=\frac{(\sqrt3-\sqrt2)}{+1} \\\\

=\sqrt3-\sqrt2\\$$

Melody
Aug 5, 2014

#5**+5 **

Okay. If you do the question first, that can help.

$$\frac{1}{\sqrt2+\sqrt3}\\\\=0.5176380902050415$$

Now, let's do C.

$$\sqrt2-\sqrt3=0.5176380902050415$$

Compare the answers.

$$0.5176380902050415=0.5176380902050415$$

So, the equation is true, and the answer is C. And yes, you can use the calculator too.

You can do it by not using the calculator.

(That helped me get the answer.)

But, there is another way, too. Do A, B, C, D, and E. That helps too, you know. That can really help.

DragonSlayer554
Aug 5, 2014