C1 Question Help.....

Could somebody please walk me though the following question and how to solve it please? I don't even know where to start.

Simplify:
5 - √3 over 2 + √3
giving your answer in the form a + b√3, where a and b are integers.

Also, if anyone knows what section of the Edexcel C1 book this topic is in, please let me know so I can find it :smile:

Thanks!

Reply 1

Zen-Ali

Multiply through the entire fraction by the conjugate of the denominator. I.e., multiply both the numerator and denominator by

2 - \sqrt3.

Reply 2

Penguins&Pandas

have you tried rationalising the denominator?

(edited 10 years ago)

Reply 3

MidnightDream

Rationalise and then simplify :smile:

Reply 4

alow

Original post by thomash131

Thanks!

\dfrac{5- \sqrt{3}}{2+ \sqrt{3}}

?

You need to rationalise the denominator.

We do this by multiplying both the numerator and the denominator by the conjugate of the denominator, as

(a- \sqrt{b})(a+ \sqrt{b}) = a^2 - b

Reply 5

Munrot07

Original post by thomash131

Thanks!

Have you heard about rationalising the denominator? I believe it is in chapter one :smile:

Reply 6

Moonwalker39

Oh god! I feel so stupid!

I literally thought to myself, could it be rationalizing the denominator? but brushed it aside because I got thrown off by the 2+ and 5-.

I'll definitely go over it now. Thanks everyone!

Reply 7

Moonwalker39

Original post by Zen-Ali

Multiply through the entire fraction by the conjugate of the denominator. I.e., multiply both the numerator and denominator by

2 - \sqrt3.

Original post by alow

\dfrac{5- \sqrt{3}}{2+ \sqrt{3}}

?

You need to rationalise the denominator.

We do this by multiplying both the numerator and the denominator by the conjugate of the denominator, as

(a- \sqrt{b})(a+ \sqrt{b}) = a^2 - b

Is this correct?

\frac{5-\sqrt 3}{2+\sqrt3}

\frac{2-\sqrt 3}{2-\sqrt3}

\frac{10-5\sqrt 3 -2\sqrt 3 + 3}{4-3}

\frac{10-7\sqrt3 + 3}{1}

13-7\sqrt 3

Thanks again everyone.

(edited 10 years ago)

Reply 8

alow

Original post by thomash131

Is this correct?

\frac{5-\sqrt 3}{2+\sqrt3}

\frac{2-\sqrt 3}{2-\sqrt3}

\frac{10-5\sqrt 3 -2\sqrt 3 + 3}{4-3}

\frac{10-7\sqrt3 + 3}{1}

13-7\sqrt 3

Thanks again everyone.

Looks good to me :smile:

Reply 9

Moonwalker39

Original post by alow

Looks good to me :smile:

YES! You don't understand how happy this makes me.

All hope for the Core 1 exam is not lost! Thank you again!

Reply 10

Zen-Ali

Original post by thomash131

YES! You don't understand how happy this makes me.

All hope for the Core 1 exam is not lost! Thank you again!

[Indent]Your answer is correct! If ever you're stuck, you can use this site to check. [X] I would only recommend using it if you've tried several things first; as a last resort.

Best of luck.[/Indent]

Reply 11

Moonwalker39

Original post by Zen-Ali

Your answer is correct! If ever you're stuck, you can use this site to check. [X] I would only recommend using it if you've tried several things first; as a last resort.

Best of luck.

Awesome! Thanks for your help.

Reply 12

Zakee

Original post by Zen-Ali

Multiply through the entire fraction by the conjugate of the denominator. I.e., multiply both the numerator and denominator by

2 - \sqrt3.

Are you Ramanujan?

Reply 13

Zen-Ali

Original post by Zakee

Are you Ramanujan?

[Indent]Lmfao, nobody on TSR is even close to Ramanujan.[/Indent]

Reply 14

Zakee

Original post by Zen-Ali

Lmfao, nobody on TSR is even close to Ramanujan.

Lol you don't understand. No Mathematician alive today is even 1% of Ramanujan. Btw Ramanujan believed in re-incarnation so technically after 100 years someone might be the next Rammy-J. :--)

Reply 15

Zen-Ali

Original post by Zakee

Lol you don't understand. No Mathematician alive today is even 1% of Ramanujan. Btw Ramanujan believed in re-incarnation so technically after 100 years someone might be the next Rammy-J. :--)