Having trouble simplifying conjugate

In summary, the conversation is about simplifying the expression (4[(SQRT(x+2)) – (SQRT2))]/x and finding the limit of this expression as x approaches 0. The first person suggests multiplying by the conjugate, but the other person explains that this takes the expression in the opposite direction of simplification. They then discuss using analytic methods to find the limit and plugging in x=0 to get the result of SQRT(2). The conversation ends with a suggestion to keep all related questions in one thread for easier tracking and help.
  • #1
ladyrae
32
0
I am having trouble simplifying

(4[(SQRT(x+2)) – (SQRT2))]/x

I multiply the conjugate and I come up with 4/((SQRT(x+2)) + SQRT2)

I'm not sure if I'm on the right track
 
Mathematics news on Phys.org
  • #2
Quite frankly, by the concept of "simplified" that I am familiar with, the only simplification of the original that is possible is simply to multiply the 4 through:

[4*sqrt(x+2) - 4*sqrt(2)]/x

"Simplified" usually means rational denominators - your multiplying by the conjugate took you in the opposite direction.

What is this for? If the problem is just to simplify the original expression, then what I gave is as far as you can go. But if you need to do something else afterward, then this "simplified" form may not be the easiest to work with. It depends on the application.
 
  • #3
Actually its a limit problem I'm working on I'm having trouble with algebra.

Find the limit by analytic methods:

lim x->0 (4[(SQRT(x+2)) – (SQRT2))]/x

The first part of the problem asked me to estimate the limit by using a table and I came up with 1.414.

I am trying to multiple the conjugate...
 
  • #4
ladyrae said:
I am having trouble simplifying

(4[(SQRT(x+2)) – (SQRT2))]/x

I multiply the conjugate and I come up with 4/((SQRT(x+2)) + SQRT2)

I'm not sure if I'm on the right track

Actually you've already done all the hard work. Notice that the form you now have, 4/((SQRT(x+2)) + SQRT2), no longer gives you 0/0 if you evaluate it at x=0. So you can just plug in x=0 and get the result of the limit. You'll get 4/(2*SQRT(2))=2/SQRT(2)=SQRT(2).
 
Last edited:
  • #5
Did you multiply (sqrt (x+2) + sqrt (2) ) to your numerator and your denominator?
You should end up with 4(x) / [x((SQRT[x+2]) + (SQRT[2]))]
cancel an x on top and bottom. now you can plug in 0s anywhere x is (only one place)and you get
4/ (SQRT[0+2] + SQRT[2])
now what have you got?
 
  • #6
whoops! mc beat me to it!
 
  • #7
Master coda is right. Your derivation was correct, and in this form the limit is trivial.

(An excellent example of where "simplified" is not the form you want.)
 
  • #8
ladyrae, why didn't you post this in your earlier limit thread, which you already created to get help with this problem? It's much easier to keep track of your questions if you keep them in one place.
 
  • #9
sorry

sorry, i thought it would be the wrong section for an algebra question.
 
  • #10
Well, it was related to your original question about the limit, so I think it's fine to put it in the same thread. Any given problem might have parts that relate to physics, calculus, and algebra, but it's easier to follow and help you with your work if it's all together in one section.
 

1. What is a conjugate?

A conjugate is a pair of numbers or expressions that are related to each other through a specific mathematical operation. In the case of simplifying conjugates, we are looking at pairs of complex numbers in the form a + bi and a - bi, where a and b are real numbers and i is the imaginary unit.

2. Why is it important to simplify conjugates?

Simplifying conjugates can make complex calculations and equations easier to work with. By simplifying, we can eliminate the imaginary unit and end up with a real number, making the problem more manageable.

3. How do I simplify conjugates?

To simplify conjugates, we use the difference of squares formula: (a + bi)(a - bi) = a2 - b2i2 = a2 + b2. This formula allows us to eliminate the imaginary unit and end up with a real number.

4. What are some common mistakes when simplifying conjugates?

One common mistake is forgetting to apply the difference of squares formula and trying to distribute the imaginary unit. Another mistake is incorrectly simplifying the conjugate, resulting in an incorrect answer.

5. Can simplifying conjugates be used in real-life applications?

Yes, simplifying conjugates is useful in various fields such as engineering, physics, and economics. It can help simplify complex calculations and equations, making them easier to solve and analyze.

Similar threads

MHB 5.12 synthetic division
    • General Math
Replies
2
Views
803
MHB [ASK] Exponents and Roots Simplification problem
    • General Math
Replies
12
Views
2K
A Simplifying a double summation
    • General Math
Replies
3
Views
1K
MHB [ASK] Equation of a Circle
    • General Math
Replies
4
Views
1K
I Simplify Tricky Equation for Purely Imaginary C with Complex Constants
    • General Math
Replies
5
Views
716
MHB Help Needed: Solving x and Simplifying Equations
    • General Math
Replies
5
Views
1K
Find the supremum of ##Y## if it exists. Justify your answer.
    • Calculus and Beyond Homework Help
Replies
20
Views
413
B Could someone verify how this simplifies?
    • General Math
Replies
15
Views
1K
MHB 8.aux.27 Simplify the trig expression
    • General Math
Replies
3
Views
713
B Simplifying the factors of a complex number's imaginary part
    • General Math
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top