Question

Help with question 5! http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w06_qp_3.pdf

Answer 1

5i or 5ii ?

Answer 2

5i first. :)

Answer 3

5i looks pretty straightforward, all you need to do is simply "rationalize" the denominator by multiplying by its "conjugate"

Answer 4

Also recall below identity :
\[(a+b)(a-b) = a^2-b^2\]

Answer 5

I'm worried about the squareroots.

Answer 6

squareroots get killed by squares, so start with the given expression :
\[\dfrac{1}{\sqrt{1+x}+\sqrt{1-x}}\]

multiply that by a special kind of 1 which was tailormade to handle squareroots: \(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\)

Answer 7

\[\begin{align}\dfrac{1}{\sqrt{1+x}+\sqrt{1-x}} ~&= ~\dfrac{1}{\sqrt{1+x}+\sqrt{1-x}}\times1\\~\\
&=\dfrac{1}{\sqrt{1+x}+\sqrt{1-x}}\times \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\end{align}\]

Answer 8

Oh, wait, I should start off with the fraction on the main equation?

Answer 9

good point ! yes i think we better start with the given expression and deduce whatever was asked. please disregard everythign above

Answer 10

Okay. So uh, how do i work that expression out, :(

Answer 11

so you want to start wid below expression circled inr ed

Answer 12

yesss.

Answer 13

\[\begin{align}

\left(\color{blue}{\sqrt{1+x}}+\color{purple}{\sqrt{1-x}}\right)
\left(\color{blue}{\sqrt{1+x}}-\color{purple}{\sqrt{1-x}}\right)

\end{align}\]


does that expression look similar to the identity I asked you to recall earlier ?

Answer 14

yes, sort of.

Answer 15

simply apply the identity :

\[\begin{align}

\left(\color{blue}{\sqrt{1+x}}+\color{purple}{\sqrt{1-x}}\right)
\left(\color{blue}{\sqrt{1+x}}-\color{purple}{\sqrt{1-x}}\right)
&= \left(\color{blue}{\sqrt{1+x}}\right)^2 -
\left(\color{purple}{\sqrt{1-x}}\right)^2\\~\\
&= (\color{blue}{1+x}) - (\color{purple}{1-x})\\~\\
&= ?
\end{align}\]

Answer 16

2X?

Answer 17

Yep!

simply apply the identity :

\[\begin{align}

\left(\color{blue}{\sqrt{1+x}}+\color{purple}{\sqrt{1-x}}\right)
\left(\color{blue}{\sqrt{1+x}}-\color{purple}{\sqrt{1-x}}\right)
&= \left(\color{blue}{\sqrt{1+x}}\right)^2 -
\left(\color{purple}{\sqrt{1-x}}\right)^2\\~\\
&= (\color{blue}{1+x}) - (\color{purple}{1-x})\\~\\
&= 2x
\end{align}\]

Answer 18

so we have :
\[\begin{align}

\left(\color{blue}{\sqrt{1+x}}+\color{purple}{\sqrt{1-x}}\right)
\left(\color{blue}{\sqrt{1+x}}-\color{purple}{\sqrt{1-x}}\right)

&= 2x
\end{align}\]

see if you can deduce the equation in 5i from this

Answer 19

I see the 2x there, but i dont really get what they want. D:

Answer 20

Wait, let me try. I think i sort of get it alrd.

Answer 21

good take ur time :)

Answer 22

Got it! Many thankss! :)))