Jump to content

How do YOU solve this equation?


Genady

Recommended Posts

Spoiler

Method 1:

4 and 1/4 can be spotted directly, as 4+1/4 = 1/4+1/(1/4), and it's clearly equivalent to a quadratic equation (once zero is ruled out as a solution), so those are the only roots.

Method 2:

I rule out x=0, as it is not a solution. Then I mulltiply by x to obtain the quadratic eq. 4x2-17x+4=0, which gives x=4 and x=1/4 via the algorithm for solving quadratic equations.

Spotting obvious solutions first can sometimes be very helpful to obtain the non-obvious ones when the equation is polynomial and has degree > 4.

Example: x5=x

What would you do?

Link to comment
Share on other sites

11 minutes ago, sethoflagos said:
  Reveal hidden contents

x=4 is one obvious root just on inspection, but if we ignore that:

Multiply both sides by 4x to give 4x2 + 4 = 17x or 4x2 - 17x+ 4 = 0

By formula x =(17 +/- (172 - 43)1/2)/8

Ans x = 4 or 1/4

Ooh! That's neat!

... what else could you be looking for?

Spoiler

Many go straight to 17/4 and miss the first approach. OTOH, some see both answers by the first approach.

 

Edited by Genady
Link to comment
Share on other sites

Spoiler

For me, one solution popped out by seeing that 4 1/4 can be said as "4 and one fourth."  So then x + 1/x is obviously 4. 

Then say the right hand part backwards, "one fourth and four" and you get the other solution, 1/4.

It's really just noticing the RH part is just a number and its reciprocal.  

 

31 minutes ago, sethoflagos said:

what else could you be looking for?

You did it the hard way.  🙂

Link to comment
Share on other sites

5 hours ago, sethoflagos said:
  Reveal hidden contents

x=4 is one obvious root just on inspection, but if we ignore that:

Multiply both sides by 4x to give 4x2 + 4 = 17x or 4x2 - 17x+ 4 = 0

By formula x =(17 +/- (172 - 43)1/2)/8

Ans x = 4 or 1/4

Ooh! That's neat!

... what else could you be looking for?

I just thought of something else, for a case when one sees one solution but not the other:

Spoiler

Instead of solving the quadratic equation, one can notice that after multiplying by x the quadratic equation will have a constant term 1, which means that the product of the roots is 1, which gives the second root, 1/4.

 

Link to comment
Share on other sites

2 hours ago, Genady said:

I just thought of something else, for a case when one sees one solution but not the other:

  Reveal hidden contents

Instead of solving the quadratic equation, one can notice that after multiplying by x the quadratic equation will have a constant term 1, which means that the product of the roots is 1, which gives the second root, 1/4.

 

My impression of equations of that type are dominated by:

x - 1/x = 1

Link to comment
Share on other sites

9 hours ago, Genady said:

Could you elabotate?

Spoiler

Your OP equation is one of a family whose roots are the positive inverse of each other.

Conversely mine is one of a family whose roots have a negative inverse relationship. 

eg. x - 1/x = 3 3/4 solves to x = 4, -1/4

x - 1/x = 1 solves to the golden ratio (as I'm sure you know) and its negative inverse. A deep, deep rabbit hole I've ventured into on occasion.

 

Compare and contrast:

x + 1/x = 51/2

Edited by sethoflagos
Link to comment
Share on other sites

1 hour ago, sethoflagos said:
  Reveal hidden contents

Your OP equation is one of a family whose roots are the positive inverse of each other.

Conversely mine is one of a family whose roots have a negative inverse relationship. 

eg. x - 1/x = 3 3/4 solves to x = 4, -1/4

x - 1/x = 1 solves to the golden ratio (as I'm sure you know) and its negative inverse. A deep, deep rabbit hole I've ventured into on occasion.

 

Compare and contrast:

x + 1/x = 51/2

I see.

A general form

Spoiler

\[x+\frac{a}{x}=b\]

has roots with the product \(a\) and the sum \(b\)

 

(However, the OP was about how people approach that equation rather than a mathematics behind it.)

(I see how YOU approach it 🙂 )

Edited by Genady
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.