How Do You Rationalize the Denominator to Solve for a+b?

  • Thread starter Thread starter terryds
  • Start date Start date
  • Tags Tags
    Complex Root
AI Thread Summary
To solve the equation √((10+4√6)/(10-4√6)) = a + b√6, the denominator must be rationalized. After rationalizing, the expression simplifies to √(49 + 20√6). By equating the resulting form to a + b√6, it is determined that a = 5 and b = 2. Therefore, a + b equals 7. The final answer is 7.
terryds
Messages
392
Reaction score
13

Homework Statement



If ##\sqrt{\frac{10+4\sqrt{6}}{10-4\sqrt{6}}}=a+b\sqrt{6}##, then a+b is ?

A) 8
B) 7
C) 6
D) 5
E) 4

The Attempt at a Solution


[/B]
This is my attempt:
##\sqrt{\frac{10+4\sqrt{6}}{10-4\sqrt{6}}}=\sqrt{\frac{10+4\sqrt{6}*(10+4\sqrt{6})}{10-4\sqrt{6}*(10+4\sqrt{6})}}=\sqrt{\frac{196+80\sqrt{6}}{4}}=\sqrt{49+20\sqrt{6}}##

Then, I got stuck.. I have no idea how to convert to form a+b√6
Please help..
 
Physics news on Phys.org
terryds said:

Homework Statement



If ##\sqrt{\frac{10+4\sqrt{6}}{10-4\sqrt{6}}}=a+b\sqrt{6}##, then a+b is ?

A) 8
B) 7
C) 6
D) 5
E) 4

The Attempt at a Solution


[/B]
This is my attempt:
##\sqrt{\frac{10+4\sqrt{6}}{10-4\sqrt{6}}}=\sqrt{\frac{10+4\sqrt{6}*(10+4\sqrt{6})}{10-4\sqrt{6}*(10+4\sqrt{6})}}=\sqrt{\frac{196+80\sqrt{6}}{4}}=\sqrt{49+20\sqrt{6}}##

Then, I got stuck.. I have no idea how to convert to form a+b√6
Please help..
Square both sides of your original equation and then rationalize the denominator on the left side.
You should get something like m + n√(6) on one side and r + s√(6) on the other side. You can equate m with r and n with s to get two equations involving a and b.
 
  • Like
Likes terryds
terryds said:
This is my attempt:
##\sqrt{\frac{10+4\sqrt{6}}{10-4\sqrt{6}}}=\sqrt{\frac{10+4\sqrt{6}*(10+4\sqrt{6})}{10-4\sqrt{6}*(10+4\sqrt{6})}}=\sqrt{\frac{196+80\sqrt{6}}{4}}=\sqrt{49+20\sqrt{6}}##

.
You have to use parentheses. The correct form is ##\sqrt{\frac{10+4\sqrt{6}}{10-4\sqrt{6}}}=\sqrt{\frac{(10+4\sqrt{6})*(10+4\sqrt{6})}{(10-4\sqrt{6})*(10+4\sqrt{6})}}##
Is not the numerator the square of something? What is its square root?
 
  • Like
Likes terryds and SammyS
Mark44 said:

Square both sides of your original equation and then rationalize the denominator on the left side.
You should get something like m + n√(6) on one side and r + s√(6) on the other side. You can equate m with r and n with s to get two equations involving a and b.
Thanks.. a is 5 and b is 2 , so a+b is 7..
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top