Question

ALGEBRA 2 HELP (will give a medal)

Answer 1

what is the simplest form of the expression sqrt3 - sqrt6 / sqrt3 + sqrt6

Answer 2

They gave you
\[\Large \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}}\]
right?

Answer 3

yes. i have answer choices too, if that helps

Answer 4

the given denominator is \(\Large \sqrt{3}+\sqrt{6}\)
the conjugate is \(\Large \sqrt{3}-\sqrt{6}\)

Notice how the + changed to -

Multiply top and bottom by the conjugate of the denominator to get...

\[\Large \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}}\]
\[\Large \frac{(\sqrt{3}-\sqrt{6}){\color{red}{*(\sqrt{3}-\sqrt{6})}}}{(\sqrt{3}+\sqrt{6}){\color{red}{*(\sqrt{3}-\sqrt{6})}}}\]

In the denominator, you can use the rule (a-b)(a+b) = a^2 - b^2 to take a shortcut and you'll notice the radicals go away in the denominator

I'll let you finish up

Answer 5

So would I do (sqrt3-sqrt6)(sqrt3+sqrt6) = 3^2 - 6^2 ?

Answer 6

close

Answer 7

you square "sqrt(3)" to get 3
you square "sqrt(6)" to get 6

ie

\[\Large (\sqrt{3})^2-(\sqrt{6})^2 = 3 - 6 = -3\]

Answer 8

the square and square root are opposite operations
they undo and cancel each other out

Answer 9

you now have

\[\Large \frac{(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6})}{-3}\]

Answer 10

see what happens when you expand out the numerator

Answer 11

okay, would i multiply (sqrt3 - sqrt6)(sqrt3 - sqrt6)?

Answer 12

yes, multiply that out

Answer 13

do i want to keep it as a square root?

Answer 14

what do you mean?

Answer 15

never mind. i put it in my calculator and wound up with 0.514718625761. but i dont think that's right