Practice problems. I am studying for an upcoming test.?

Question

I want to make sure I understand how to solve this problem. Here is what I have so far.

3 sqrt2/2 sqrt6-sqrt10

From here I know that I have to multiply by the top and bottom of the problem again on the right side of the problem

So now I have 3 sqrt2/2 sqrt6-10*3 sqrt2/2 sqrt6-10

Will I have 9 sqrt2 now? How do I multiply the bottom? Will the denominator be 2?

Answer 1

Thanks for the clarification in your e-mail... I see that the denominator should include the subtracted term:

...... 3 sqrt(2)
------------------------
2 sqrt(6) - sqrt(10)

When you have this situation, you need to create a difference of squares in the denominator. In other words if you have sqrt(a) - sqrt(b), multiply by sqrt(a) + sqrt(b). This will make a difference of squares and get rid of the square roots.

In your equation, multiply the numerator and denominator by 2 sqrt(6) + sqrt(10)

...... 3 sqrt(2) [2 sqrt(6) + sqrt(10) ]
-------------------------- ---------------------------
[2 sqrt(6) - sqrt(10)] [2 sqrt(6) + sqrt(10) ]

Simplify the denominator:
3 sqrt(2) [2 sqrt(6) + sqrt(10) ]
------------------- --------------------
..... 2 sqrt(6)² - sqrt(10)²

3 sqrt(2) [2 sqrt(6) + sqrt(10) ]
------------------ ---------------------
............... 4 * 6 - 10

3 sqrt(2) [2 sqrt(6) + sqrt(10) ]
------------------- --------------------
................... 14

Now multiply through in the numerator:
6 sqrt(12) + 3 sqrt(20)
-----------------------------
...............14

Simplify the radicals:
6 * 2 sqrt(3) + 3 * 2 * sqrt(5)
------------------ -------------------
................... 14

And the final answer is:

12 sqrt(3) + 6 sqrt(5)
----------------------------
.............. 14

Answer 2

3rt2/(2rt6-rt10)
rationalise the Dr by multiplyingthe Ner and Dr by 2rt6+rt10
3rt2(2rt6+rt10)/24-10
=3rt2(2rt6+rt10)/14