Question

Rationalize the numerator.
a)\[\sqrt{2+h}+\sqrt{2-h}/h\]
b)\[xsqrt{x}-8/x-4\]

Answer 1

the key thing is to use the below identiy cleverly :
\[(a+b)(a-b) = a^2 - b^2\]

Answer 2

how???

Answer 3

do you know what the "conjugate" is?

Answer 4

yes

Answer 5

multiply by the conjugate

Answer 6

i tried, it's not working...

Answer 7

i.e. multiplly
\[\frac{\sqrt{2+h}+\sqrt{2-h}}{h}\times \frac{\sqrt{2+h}-\sqrt{2-h}}{\sqrt{2+h}-\sqrt{2-h}}\]

Answer 8

what do you get in the numerator?

Answer 9

1

Answer 10

i need help... :( @ganeshie8

Answer 11

I don't get 1 for the numerator:\[(2+h) - (2-h)\]

Answer 12

Is \[2+h-2+h\]

Answer 13

Which is 2h.

Answer 14

Does that make sense?

Answer 15

gotcha, i made stupid mistakes. Thanks for pointing that out! :)

Answer 16

No worries :)

Answer 17

do you know how to do b)?

Answer 18

Is b \[\frac{ x \sqrt{x-8} }{ x-4 }\]

Answer 19

?

Answer 20

no, it's
\[\frac{ x \sqrt{x}-8 }{ x-4 }\]
Sorry of the confusion...

Answer 21

Ah, gotcha. No worries.

Answer 22

Let me work it out, just a sec.

Answer 23

okay thanks!

Answer 24

This is another multiply by the conjugate problem.

Answer 25

is this correct??
\[\frac{ x^2+4x+16 }{ x \sqrt{x}+8 }\]

Answer 26

Hm, that's not what I get.

Answer 27

Did you do it like this?
\[\frac{ x \sqrt{x}-8 }{ x-4 }\times \frac{ x \sqrt{x}+8 }{ x \sqrt{x}+8 }\]

Answer 28

yes

Answer 29

then i got
\[\frac{ x^3-64 }{ x-4(x \sqrt{x}-8 }\]

Answer 30

:) That's almost what I get. I just have a + before the 8 in the denominator.

Answer 31

i meant +8)*

Answer 32

Ok, cool! :D

Answer 33

we can reduce that, right?

Answer 34

Yep.

Answer 35

Do you know how?

Answer 36

well, i got
\[\frac{ (x-4)(x^2+4x+16) }{ (x-4)(x \sqrt{x}+8) }\]
than I reduce the (x-4)

Answer 37

I also got :\[\frac{x^3-64}{(x-4)(x\sqrt{x}+8)}\]x\(^3\)-64 is a difference of cubes. \(a^3-b^3 = (a-b)(a^2+ab+b^2)\)
This correlates to : \((x)^3 -(4)^3\)

Answer 38

Perfect!

Answer 39

Whoah, faster than me!

Answer 40

thanks for helping guys! :D @JoannaBlackwelder @Jhannybean

Answer 41

No problem :) Don't forget to reduce the like terms in both numerator and denominator!

Answer 42

You're welcome! :)