1. ## surds iii

I got 1/6 for (a) but (b) is beyond me. Thanks.

2. 1/6 is right.

For b, you just want to show that the left is equal to the right. We do this by substituting and simplifying as follows:

$\displaystyle (r-p)^2=9-6\sqrt{2}$
$\displaystyle (\sqrt{6}-\sqrt{3})^2 \Rightarrow (\sqrt{6})^2-2\sqrt{6}\sqrt{3}+(\sqrt{3})^2 \Rightarrow 6-2\cdot 3\sqrt{2}+3 \Rightarrow 9-6\sqrt{2}$ which was to be shown.

I got 1/6 for (a) but (b) is beyond me. Thanks.
(b) is actually not that hard, what did you try?

begin with $\displaystyle (r - p)^2$

$\displaystyle (r - p)^2 = r^2 - 2rp + p^2$

$\displaystyle = ( \sqrt {6} )^2 - 2 \sqrt {6} \sqrt {3} + ( \sqrt {3} )^2$

Can you take it from here?

4. This is what I did:

(sq.rt6 - sq.rt3) (sq.rt6 - sq.rt3) = 9 - 6 sq.rt2

(sq.rt6)^2 - sq.rt18 - sq.rt18 + sq.rt9

This is what I did:

(sq.rt6 - sq.rt3) (sq.rt6 - sq.rt3) = 9 - 6 sq.rt2

(sq.rt6)^2 - sq.rt18 - sq.rt18 + sq.rt9

why?

(sqrt(6))^2 = 6

sqrt(9) = 3

-sqrt(18) - sqrt(18) = -2sqrt(18) = -2sqrt(9*2) = -2sqrt(9)*sqrt(2) = -2(3)sqrt(2) = -6sqrt(2)

put those all together and you have the answer

This is what I did:

(sq.rt6 - sq.rt3) (sq.rt6 - sq.rt3) = 9 - 6 sq.rt2

(sq.rt6)^2 - sq.rt18 - sq.rt18 + sq.rt9

$\displaystyle (\sqrt{6} - \sqrt{3})(\sqrt{6} - \sqrt{3})$

$\displaystyle = (\sqrt{6})^2 - \sqrt{18} - \sqrt{18} + \sqrt{9}$

$\displaystyle = 6 - 2\sqrt{18} + 3$

$\displaystyle = 9 - 2\sqrt{18}$

$\displaystyle = 9 - 2\sqrt{9 \cdot 2}$

$\displaystyle = 9 - 2\sqrt{3^2 \cdot 2}$

$\displaystyle = 9 - 2 \cdot3 \sqrt{2}$

$\displaystyle = 9 - 6\sqrt{2}$

-Dan

7. 6 - 2 sq.rt18 + 3

9 - 2 sq.rt 18

sq.rt18 = sq.rt9 * sq.rt 2 = 3 sq.rt2

but that gives 9 - 3 sq.rt2

EDit: thanks rualin, 9 - sq.rt 2 * 3 sq.rt2

9 - 6 sq.rt2

8. Times the 2 that was already there = 6 sqrt(2)

9. By the way, what's a surd? Seems like absurd...making no sense? Excuse my ignorance.

10. Originally Posted by rualin
By the way, what's a surd? Seems like absurd...making no sense? Excuse my ignorance.
My first answer to a question.......

A surd is, of course, a verb. It's the past tense of 'to sur' as in the make someone a knight, like what the queen does (here in the Uk anyway).

11. Originally Posted by rualin
By the way, what's a surd? Seems like absurd...making no sense? Excuse my ignorance.
a surd is an unresolved radical expression.

example, $\displaystyle \sqrt {9}$ is a surd, since we could simplify (or resolve) this to get 3, but we leave it in its radical form because it's convenient to do so in our problem. that's a surd

12. Ahhh! That makes a lot more sense. Thanks