Solving Equation Problems w/ Componendo & Dividendo

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In summary, the conversation is about a math problem involving a pure quadratic equation. The problem is solved until the equation (5x^2 + 7)/(5x^2 - 7) = 16/9, where the confusion arises with the use of "componendo and dividendo." The person asking for help is also confused by another example involving the equation √(3x^2 - 7x - 30) - √(2x^2 - 7x - 5) = x - 5 and the incorrect equation (3x^2 - 7x - 30)^2 - (2x^2 - 7x - 5)^2 = (x
  • #1
1/2"
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Hey this is worked out sum from my book and I am having a bit problem in understanding it

Homework Statement


(It's a pure quadratic equation)
If (35-2x)/9+(5x 2+7)/(5x 2-7)=(17-2/3x)/3 find x

Well its worked out well till
(5x 2+7)/(5x 2-7)=16/9
and the comes the real confusion is
Its given
5x 2/7= (16+9)/(16-9)=25/7 [componendo and dividendo]
What is this" componendo and dividendo"?:rolleyes::confused:
I don't know !Please help!:frown:

Another e.g. is
√( 3x^2-7x-30)-√(2x^2-7x-5)=x-5
then they write
Identically
(3x^2-7x-30)-2x^2-7x-5)=x^2-25(!?)
Is this also by the same" componendo and dividendo"?:confused:
Plz help me out! I really don't this componendo and dividendo !
 
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  • #2
1/2" said:
Hey this is worked out sum from my book and I am having a bit problem in understanding it

Homework Statement


(It's a pure quadratic equation)
If (35-2x)/9+(5x 2+7)/(5x 2-7)=(17-2/3x)/3 find x

Well its worked out well till
(5x 2+7)/(5x 2-7)=16/9
and the comes the real confusion is
Its given
5x 2/7= (16+9)/(16-9)=25/7 [componendo and dividendo]
What is this" componendo and dividendo"?:rolleyes::confused:
I don't know !Please help!:frown:
It means composing and dividing. Are you studying a math book in Spanish?

The equation below is fine.
(5x 2+7)/(5x 2-7)=16/9
I don't follow what you have in the equation below that one, but if you continue working on the equation above you should be able to arrive at the solution(s). Multiply both sides by 9(5x2 -7).


1/2" said:
Another e.g. is
√( 3x^2-7x-30)-√(2x^2-7x-5)=x-5
then they write
Identically
(3x^2-7x-30)-2x^2-7x-5)=x^2-25(!?)
Is this also by the same" componendo and dividendo"?:confused:
Plz help me out! I really don't this componendo and dividendo !
Where are you getting this stuff? Is this from some other student? Whatever, this isn't how you solve equations like this. You can't get to the 2nd equation from the first equation. It looks like whoever did this squared each term on the left side (which is not a valid operation), and then incorrectly squared each term on the right side. This is totally bogus.
 
Last edited:
  • #3
I think so too and I'm going nuts over it .
anyawy thanks A LOT!:smile:
 

FAQ: Solving Equation Problems w/ Componendo & Dividendo

1. What is the Componendo and Dividendo method for solving equations?

The Componendo and Dividendo method is a technique used to solve equations involving two variables. It involves adding and dividing both sides of the equation by a specific ratio in order to simplify and solve for the unknown variables.

2. When should I use the Componendo and Dividendo method?

The Componendo and Dividendo method is most useful for solving equations where the variables are in the form of fractions. It can also be used when there are multiple variables on both sides of the equation.

3. What is the difference between Componendo and Dividendo?

The terms "componendo" and "dividendo" refer to the actions of adding and dividing, respectively. In the Componendo and Dividendo method, these actions are used to manipulate the equation and solve for the unknown variables.

4. Can the Componendo and Dividendo method be used for all types of equations?

No, the Componendo and Dividendo method is best suited for equations with two variables and where the variables are in the form of fractions. It may not be effective for solving equations with more than two variables or those with complex algebraic expressions.

5. What are some common mistakes to avoid when using the Componendo and Dividendo method?

One common mistake is applying the Componendo and Dividendo method incorrectly, which can lead to incorrect solutions. It is important to carefully follow the steps and make sure that the ratio used for adding and dividing is applied to both sides of the equation. Another mistake is forgetting to check the solution by substituting it back into the original equation.

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