+0

# In question 21

+3
843
6

In question 21

can I solve it without calculater !! Aug 5, 2014

#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 a2 and b2 are not!

$$\frac{1}{\sqrt2+\sqrt3}\\\\ =\frac{1}{\sqrt2+\sqrt3} \times \frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}\\\\ =\frac{(\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)} \\\\ =\frac{(\sqrt2-\sqrt3)}{(\sqrt2)^2-(\sqrt3)^2} \\\\ =\frac{(\sqrt2-\sqrt3)}{2-3} \\\\ =\frac{(\sqrt2-\sqrt3)}{-1} \\\\ =\frac{(\sqrt3-\sqrt2)}{+1} \\\\ =\sqrt3-\sqrt2\\$$

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Aug 5, 2014

#1
+8

Does it involve division too?

Aug 5, 2014
#2
+3

I talk about the question !

Aug 5, 2014
#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 a2 and b2 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
#4
+3

hey melody !

you are amaiznnnnnnnnnnnnnng !!!!

thanx my friend

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.

$$\sqrt2-\sqrt3=0.5176380902050415$$

$$0.5176380902050415=0.5176380902050415$$

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.

Aug 5, 2014
#6
+8

Thanks 15x3

Flattery and thumbs up will get you a very long way with me. DS,

15x3 expressely asked for this to be solved WITHOUT a calculator.  So, did you really answer the question?

This is a rhetorical question - don't dare post an answer!

Aug 5, 2014