Question

im having a problem understanding algebra stuff. i can grasp how to multiply radicals. help anyone?

Answer 1

\[\sqrt{a}~\sqrt{b}=\sqrt{a*b}\]
not too sure how thats confusing ....

Answer 2

i mean its hold on... https://media.glynlyon.com/g_alg02_2012/7/q11174.gif

Answer 3

conjugates help with simplification

Answer 4

rationalizing the denominator

Answer 5

i get how to find the gonjugates.. maybe im just over thinking it -_-

Answer 6

change the operation, and multiply

Answer 7

(a+b) <-> (a-b)

are conjugates

Answer 8

the beauty is:\[(a+b)(a-b) = a^2 -b^2\]

Answer 9

the top will just have to be whatever it plays out as ..

Answer 10

okay

Answer 11

what i dont get is would it be 12 radical 36?

Answer 12

\[\frac{c+d}{a-b}\frac{a+b}{a+b}=\frac{ca+cb+da+db}{a^2-b^2}\]
the specific values are just distractions

Answer 13

12(6) +9sqrt(12)+20sqrt(12)+15(4)
------------------------------
16(6) - 9(4)

Answer 14

uhh, got a little carried away :) *-9(2) is that under part :)

Answer 15

and 15(2) up top ....

Answer 16

12(6) +9sqrt(12)+20sqrt(12)+15(2)
------------------------------
16(6) - 9(2)


thats better

Answer 17

*confused*

Answer 18

not too sure what the confusion it ...

Answer 19

wait.. hold on

Answer 20

so 15 radical 2 is n top and -9 radical 2 on bottom. or is that supposed to be squared

Answer 21

you recall how to "FOIL"? i dont really like that term but it seems to be what passes of math these days

Answer 22

\((3\sqrt6+5\sqrt2)\) \((4\sqrt6+3\sqrt{2})\)
------------------------
\((4\sqrt6-3\sqrt{2})\) \((4\sqrt6+3\sqrt{2})\)




\(3(4)\sqrt{6(6)}+5(4)\sqrt{2(6)}+3(3)\sqrt{6(2)}+5(3)\sqrt{2(2)}\)
---------------------------------------------
\(4(4)\sqrt{6(6)}-3(4)\sqrt{2(6)}+3(4)\sqrt{2(6)}-3(3)\sqrt{2(2)}\)




\(12\sqrt{36}+20\sqrt{12}+9\sqrt{12}+15\sqrt{4}\)
-----------------------------
\(16\sqrt{36}-9\sqrt{4}\)



\(12(6)+20\sqrt{12}+9\sqrt{12}+15(2)\)
-----------------------------
\(16(6)-9(2)\)