Solving Equation: (x^2 + 4x) / 3 + 84 / (x^2 + 4x) = 11 - Step by Step Guide

[roger]

hi ,

I wanted to find out all the possible ways to solve :

(x^2 + 4x) / 3 + 84 / (x^2 +4x) = 11


Please could you show me in each case the method of working out .


Thanks a lot.


Roger

 

[arildno]

There are infinite ways of "solving" an equation.
For example, you may always insert a "solving step" by adding a-a=0 to one side of an equation (where a is some number).

Hence, we are more interested in finding EFFICIENT ways of solving an equation, rather than POSSIBLE ways.

In your case, it seems the simplest way of solving this, is to introduce the variable:
$$y=x^{2}+4x$$
Multiplying your original equation by "y", and rearranging, gives:
$$\frac{1}{3}y^{2}-11y+84=0$$

Use this in your further calculations.

 

[The Bob]

(x^2 + 4x) / 3 + 84 / (x^2 +4x) = 11

Might be me being stupid but does this equation mean $$\frac{(x^2 + 4x)}{\frac{3 + 84}{(x^2 +4x)}} = 11$$ ?

The Bob (2004 ©)

 

[Nylex]

Might be me being stupid but does this equation mean $$\frac{(x^2 + 4x)}{\frac{3 + 84}{(x^2 +4x)}} = 11$$ ?

The Bob (2004 ©)

That's what I thought.

 

[arildno]

roger's notation is quite consistent with how you usually plug in formulas in a computer program, in which case the expression is to be read as:
$$\frac{x^{2}+4x}{3}+\frac{84}{x^{2}+4x}}=11$$

 

[roger]

roger's notation is quite consistent with how you usually plug in formulas in a computer program, in which case the expression is to be read as:
$$\frac{x^{2}+4x}{3}+\frac{84}{x^{2}+4x}}=11$$

Thanks, but is there any other method without substituting for y ?

I'm assuming substituting for y, is what you meant by...'' introduce the variable '' ?



Roger

 

[dextercioby]

Yes.That's what he meant.I hope u solved the eq.One more thing though,please learn how to edit formulas with Tex.People might get confused due to your inability of getting them clear with what u want...

Daniel.

 

[The Bob]

$$\frac{x^{2}+4x}{3}+\frac{84}{x^{2}+4x}}=11$$

=> $$\frac{[(x^2 + 4x)(x^2 + 4x)]+(84\times3)}{3x^2 + 12x}=11$$

It is another way to solve it up it would take a while and would not be benifical at all.

The Bob (2004 ©)

 

[dextercioby]

Forget it,Bob.It's 4-th order.It would take a couple of hours to solve it...

Daniel.

 

[The Bob]

Forget it,Bob.It's 4-th order.It would take a couple of hours to solve it...

I know. That is why I said it was unbenifical because it would take so long to solve.

$$\frac{x^4 + 8x^3 + 16x^2 +252}{3x^2 + 12x} = 11$$

The Bob (2004 ©)

 

[roger]

Well this has made me curious.

How would the 4th order equation be solved or any other similar equation of same or higher order ?

Roger

 

[dextercioby]

No higher order than 4 general algebraic equation can be solved.This was shown round 200 yrs ago by Abel and Ruffini.

Check out the wolfram site.Search for "quartic".

Daniel.

 

[arildno]

Another solution procedure is the guess&try method.
Occasionally, it works surprisingly well.

 

[BigStelly]

Or you could do a numeric solution with a computer... but i don't think that's how this praticular problem was meant to be solved. But its true no equation can be solved by traditional algebraic methods in orders higher than 4th.

 

[HallsofIvy]

No higher order than 4 general algebraic equation can be solved.This was shown round 200 yrs ago by Abel and Ruffini.

Check out the wolfram site.Search for "quartic".

Daniel.

There cannot exist a formula using only roots because there exist solutions to polynomial equations of degree 5 or higher that cannot be written in terms of roots.

I would not use the phrase "cannot be solved"!

 

[dextercioby]

Sorry,you're right...I went too far with the generalization and used inappropriate words.

It won't happen...

Daniel.

 

[The Bob]

$$\frac{x^4 + 8x^3 + 16x^2 +252}{3x^2 + 12x} = 11$$

$$x^4 + 8x^3 + 16x^2 + 252 = 33x^2 + 132x$$

$$f(x) = x^4 + 8x^3 - 17x^2 -132x + 252 = 0$$

f(3) = 0

f(2) = 0

$$(x - 3)(x - 2)( x^2 +13x + 42) = 0$$

Problem is that the answer to $$x^2 +13x + 42$$ is x = 6 or 7 which doesn't work so it works to a point.

The Bob (2004 ©)