Question

@ganeshie8 Last question?

Answer 1

I started off by multiply the numerator and the denominator by the denominator
So I had 9s - 18 + 9sqrt 17t, but I'm not sure about the denominator

Answer 2

we need to multiply wid conjugate of denominator

Answer 3

s + 2sqrt(17t)

Answer 4

So when you multiply the denominator by the denominator, you add?

Answer 5

no, we're multiplying cuz we want to get rid of radicals in the denominator. let me show u the full s olution. u can ask questions after that okay ?

Answer 6

Okay c:

Answer 7

\(\large \frac{9}{s-2\sqrt{17t}}\)

\(\large \frac{9}{s-2\sqrt{17t}} \times \frac{s+2\sqrt{17t}}{s+2\sqrt{17t}}\)

Answer 8

\(\large \frac{9}{s-2\sqrt{17t}}\)

\(\large \frac{9}{s-2\sqrt{17t}} \times \frac{s+2\sqrt{17t}}{s+2\sqrt{17t}}\)

\(\large \frac{9(s+2\sqrt{17t})}{(s-2\sqrt{17t})(s+2\sqrt{17t})} \)

Answer 9

bottom,
apply the identity : \((a+b)(a-b) = a^2-b^2\)

Answer 10

\(\large \frac{9}{s-2\sqrt{17t}}\)

\(\large \frac{9}{s-2\sqrt{17t}} \times \frac{s+2\sqrt{17t}}{s+2\sqrt{17t}}\)

\(\large \frac{9(s+2\sqrt{17t})}{(s-2\sqrt{17t})(s+2\sqrt{17t})} \)

\(\large \frac{9(s+2\sqrt{17t})}{s^2-(2\sqrt{17t})^2} \)

\(\large \frac{9s+18\sqrt{17t}}{s^2-4 \times 17t} \)

\(\large \frac{9s+18\sqrt{17t}}{s^2-68t} \)

Answer 11

see if that makes more/less sense

Answer 12

How come we multiplied by s + 2 sqrt17t instead of s - 2 sqrt17t?

Answer 13

u tell me why multiplying by s - 2 sqrt17t wil NOT get rid of radicals in denominator ? :)

Answer 14

Umm, because it wouldn't cancel out?

Answer 15

yup, radical wont go away, if u multiply the denominator wid the same

Answer 16

in denominator u wud get :-

(a-b)(a-b) = (a-b)^2

Answer 17

which is not useful at all, since radicals will exist...

Answer 18

so keep this mind :-

if u see below :

\(\huge \frac{something}{a+b\sqrt{c}}\)

Answer 19

multiply top and bottom wid : \(\large a- b \sqrt{c}\)

that takes care of radicals in denominator

Answer 20

always multiply the OPPOSITE sign..

Answer 21

Okay, is that why it went from (2sqrt17t)^2 to -4 * 17t?

Answer 22

square and radical cancelled...

Answer 23

\(\large \frac{9}{s-2\sqrt{17t}}\)

\(\large \frac{9}{s-2\sqrt{17t}} \times \frac{s+2\sqrt{17t}}{s+2\sqrt{17t}}\)

\(\large \frac{9(s+2\sqrt{17t})}{(s-2\sqrt{17t})(s+2\sqrt{17t})} \)

\(\large \frac{9(s+2\sqrt{17t})}{s^2-(2\sqrt{17t})^2} \)

\(\large \frac{9s+18\sqrt{17t}}{s^2-4 \times 17t} \) <<<<<<<<<

\(\large \frac{9s+18\sqrt{17t}}{s^2-68t} \)

Answer 24

If the square and radical canceled, how did 2 still become 4?

Answer 25

good question :)
cuz, 2 is outside the radical :)

Answer 26

\(\large \frac{9}{s-2\sqrt{17t}}\)

\(\large \frac{9}{s-2\sqrt{17t}} \times \frac{s+2\sqrt{17t}}{s+2\sqrt{17t}}\)

\(\large \frac{9(s+2\sqrt{17t})}{(s-2\sqrt{17t})(s+2\sqrt{17t})} \)

\(\large \frac{9(s+2\sqrt{17t})}{s^2-(2\sqrt{17t})^2} \) <<<<<<<

\(\large \frac{9s+18\sqrt{17t}}{s^2-4 \times 17t} \)

\(\large \frac{9s+18\sqrt{17t}}{s^2-68t} \)

Answer 27

ohh, that makes sense okay:D I get it:3

Answer 28

I am so thankful that this is the last question because my brain is fried xD

Answer 29

it deserves some rest after the long day of hardwork im sure :)

Answer 30

Haha indeed, rest from math anyway, I have history and English assignmetns to complete, then it will get its rest:3

Answer 31

OMG! have fun :))

Answer 32

Thanks! Thank you for your help!

Answer 33

I appreciate it, I wouldn't have understood it nor completed it as quickly without your tutoring!