Question

Algebra 2 help?
MEDAL + FAN

Rationalize the denominator and simplify. 12/√7+2

Answer 1

multiply the system by the complex conjugate
\[\sqrt{7}-2/(\sqrt{7}-2)\]

Answer 2

multiply nr and dnr with sqrt(7)-2

Answer 3

\[\frac{ 12 }{ \sqrt{7}+2 }\frac{ \sqrt{7}-2 }{ \sqrt{7}-2 }\]

Answer 4

\[ \frac{12}{\sqrt{7}+2}=\frac{12}{\sqrt{7}+2}\frac{\sqrt{7}-2}{\sqrt{7}-2}=\frac{12\sqrt{7}-24}{7-4}=4\sqrt{7}-8\]

Answer 5

and it is not called the complex conjugate. Just the conjugate in this case

Answer 6

Okay, I think I see it, but can you help me with another just so I can confirm that I understand how to do it? @kirbykirby

Answer 7

ok sure

Answer 8

6/√2-√3

Answer 9

So this is similar, you have to multiply the top and bottom by the conjugate. The definition of the conjugate is:
If you have \(a + b\), the conjugate is \(a - b\)
If you have \(a-b\), then conjugate is \(a+b\) (you just change the middle sign)

So,
\[ \frac{6}{\sqrt{2}-\sqrt{3}}=\frac{6}{\sqrt{2}-\sqrt{3}}\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}\]

Answer 10

Okay so that then turns to 6√2+6√3/√4-√3?

Answer 11

the numerator is correct, but the denominator isn't

Answer 12

Oops. I meant √4-√6. Is that right? Or would it be √4-√9?

Answer 13

\(-\sqrt{9}\)
since: \(\sqrt{3}\sqrt{3}=\sqrt{3\times 3}=\sqrt{9}\)

Answer 14

Alright. And then after that is where I get stuck because I don't understand how to simplify it.

Answer 15

so you have in the denominator:
\(\sqrt{4}-\sqrt{9}=?\) can you take the square roots of these numbers?

Answer 16

Would I just subtract 9 from 4?

Answer 17

@kirbykirby

Answer 18

\(\sqrt{4}=2\)
\(\sqrt{9}=3\)

Since \(2^2=4\) and \(3^2=9\)

Answer 19

Ohhh! I see now.

Answer 20

What do I do next? @@kirbykirby

Answer 21

so in the denominator, 2 - 3 = -1
soo now you have \[ \frac{6\sqrt{2}+6\sqrt{3}}{-1}\]

Now, recall:
\[\frac{x+y}{a}=\frac{x}{a}+\frac{y}{a}\]

so, you can separate your fraction as:
\[ \frac{6\sqrt{2}}{-1}+\frac{6\sqrt{3}}{-1}\]