Can someone show me how this is solved algebraically

  • Thread starter chris_0101
  • Start date
In summary, the student is seeking help to show the algebraic relationship between two equations. They have tried rationalizing the denominator but have not been successful. The expert suggests multiplying out the numerator over c^2 and using the rules for square roots and fractions to simplify the equations. The student is unsure of how to multiply out the numerator and the expert explains the steps. The final simplified equation is given as (u'^2 + 2u'v + v^2)/((u'^2 v^2)/c^4) + (2((u'v)/c^2) +1.
  • #1
chris_0101
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0

Homework Statement


I need to show how the first equation gives the second equation algebraically
img1.PNG



Homework Equations



I've tried multiple methods such as rationalizing the denominator with no avail.

The Attempt at a Solution



Any help will be greatly appreciated,

Thanks.
 
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  • #2
chris_0101 said:

Homework Statement


I need to show how the first equation gives the second equation algebraically
View attachment 51672


Homework Equations



I've tried multiple methods such as rationalizing the denominator with no avail.

The Attempt at a Solution



Any help will be greatly appreciated,

Thanks.

Multiply out the numerator of that mess over c^2. Then find a quick common divisor. Then use the rules :

[itex]\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}[/itex]

and

[itex]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}[/itex]
 
  • #3
I'm not sure what you mean by "multiply out the numerator of that mess over c^2"?
 
  • #4
chris_0101 said:
I'm not sure what you mean by "multiply out the numerator of that mess over c^2"?

The quantity [itex](\frac{u'+v}{1+vu'/c^2})^2[/itex]
 
  • #5
yea, but how does one multiply out the numerator?
do you mean like this:
(a/b)/c = (a/b)*(1/c)
 
  • #6
chris_0101 said:
yea, but how does one multiply out the numerator?
do you mean like this:
(a/b)/c = (a/b)*(1/c)

...

[itex](\frac{u'+v}{1+vu'/c^2})^2 = (\frac{u'+v}{1+vu'/c^2})(\frac{u'+v}{1+vu'/c^2})[/itex]

Now some simple bedmas and you're done...
 
  • #7
This is what I got:

(u'^2 + 2u'v + v^2)/((u'^2 v^2)/c^4) + (2((u'v)/c^2) +1

Is this right?
 

1. How do I solve algebraic equations?

The first step in solving an algebraic equation is to simplify both sides by combining like terms. Then, isolate the variable by performing inverse operations on both sides. Remember to apply the same operation to both sides to maintain the equation's balance. Once the variable is isolated, solve for it by performing the necessary operations.

2. What are inverse operations?

Inverse operations are operations that "undo" each other. For example, addition and subtraction are inverse operations, as well as multiplication and division. When solving an algebraic equation, use inverse operations to isolate the variable.

3. Can you provide an example of solving an algebraic equation?

Sure! Let's say we have the equation 2x + 5 = 17. To solve for x, we first need to simplify both sides by combining like terms. In this case, we can combine 2x and 5 to get 2x + 5. Then, we can subtract 5 from both sides to isolate the variable. This gives us 2x = 12. Finally, we divide both sides by 2 to get x = 6. Therefore, the solution to the equation is x = 6.

4. Is there a specific order to solving algebraic equations?

Yes, there is a specific order to follow when solving algebraic equations. It is commonly known as the "PEMDAS" rule, which stands for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). This order ensures that the equation is solved correctly.

5. Are there any shortcuts for solving algebraic equations?

There are some shortcuts that can be used to solve algebraic equations, such as distributing and factoring. Distributing involves multiplying a number or term to each term inside parentheses, while factoring involves finding common factors and grouping them. These shortcuts can make solving equations faster and easier, but it is important to understand the concepts behind them.

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