x^5 + y^5 = ?

[nycmathguy]

I found a video on You Tube featuring this crazy question.

Find x^5 + y^5.

{x + y = 2
{xy = 3
{x^5 +y^5 = ?

Let me see.

I can solve xy = 3 for either variable.

y = 3/x

I then can plug y = 3/x into x + y = 2 to find x.

x + y = 2

x + (3/x) = 2

x^2 + 3 = 2

x^2 = 2 - 3

x^2 = -1

Stuck here. I can't take the square root of a negative number.

What would you do?

NOTE: I can watch the entire video and learn how it is done. I'd rather try to solve it on my own without the video clip. It's more fun this way.

 

[MathLover1]

x + (3/x) = 2

x^2 + 3 = 2=> how did you get this???????

and you can take the square root of a negative number, sqrt(-1)=i

 

[nycmathguy]

x + (3/x) = 2

x^2 + 3 = 2=> how did you get this???????

and you can take the square root of a negative number, sqrt(-1)=i

I got x^2 + 3 = 2 by doing this:

x + (3/x) = 2

x^2 + 3 = 2

x^2 = 2 - 3

x^2 = -1

I see my error.

x + (3/x) = 2

Multiply both sides by x to clear the fraction on the left side.

x^2 + 3 = 2x

x^2 - 2x + 3 = 0

I will take it from here.

 

[HallsofIvy]

I found a video on You Tube featuring this crazy question.

Find x^5 + y^5.

{x + y = 2
{xy = 3
{x^5 +y^5 = ?

Let me see.

I can solve xy = 3 for either variable.

y = 3/x

I then can plug y = 3/x into x + y = 2 to find x.

x + y = 2

x + (3/x) = 2

x^2 + 3 = 2

No! You multiplied the left side by x but not the right.
x^2+ 3= 2x.
x^2- 2x= -3
x^2- 2x+ 1= -2
(x- 1)^2= -2

You still have a square equal to a negative number so x is not a real number.
x- 1= +/- i sqrt(2)
x= 1+/- i sqt(2). Notice that both have absolute value sqrt(3)

Now y= 3/x.
If x= 1+ i sqrt(2)
y= 3/(1+ i sqrt(2))= 3(1- i sqrt(2))/[(1+ i sqrt(2))(1- i sqrt(2))]= 3(1- i sqrt(2))/3= 1- i sqrt(1)
so x+ y= 2.

If x= 1- i sqrt(2)
y= 3/(1- i sqrt(2))= 3(1+ i sqrt(2))/[(1- i sqrt(2))(1+ i sqrt(2))]= 3(1+ i sqrt(2))/3= 1+ i sqrt(1)
so x+ y= 2
x^2 = 2 - 3

x^2 = -1

Stuck here. I can't take the square root of a negative number.

What would you do?

NOTE: I can watch the entire video and learn how it is done. I'd rather try to solve it on my own without the video clip. It's more fun this way.[/QUOTE]

 

[nycmathguy]

No! You multiplied the left side by x but not the right.
x^2+ 3= 2x.
x^2- 2x= -3
x^2- 2x+ 1= -2
(x- 1)^2= -2

You still have a square equal to a negative number so x is not a real number.
x- 1= +/- i sqrt(2)
x= 1+/- i sqt(2). Notice that both have absolute value sqrt(3)

Now y= 3/x.
If x= 1+ i sqrt(2)
y= 3/(1+ i sqrt(2))= 3(1- i sqrt(2))/[(1+ i sqrt(2))(1- i sqrt(2))]= 3(1- i sqrt(2))/3= 1- i sqrt(1)
so x+ y= 2.

If x= 1- i sqrt(2)
y= 3/(1- i sqrt(2))= 3(1+ i sqrt(2))/[(1- i sqrt(2))(1+ i sqrt(2))]= 3(1+ i sqrt(2))/3= 1+ i sqrt(1)
so x+ y= 2
x^2 = 2 - 3

x^2 = -1

Stuck here. I can't take the square root of a negative number.

What would you do?

NOTE: I can watch the entire video and learn how it is done. I'd rather try to solve it on my own without the video clip. It's more fun this way.

[/QUOTE]

I did the work on paper but forgot where I put it. I never got a chance to upload my work.

 

[MathLover1]

Find x^5 + y^5.

x + y = 2...............eq1
xy = 3..............eq2

xy = 3..............eq2 solve for y

y = 3/x..............eq(a)

x + 3/x = 2...............eq1, solve for x
x^2/x + 3/x = 2
(x^2 + 3)/x = 2
x^2 + 3 = 2x

x^2 -2x+3=0........using quadratic formula you get

x = 1 +i*sqrt(2)
or
x = 1 - i*sqrt(2)

y = 3/x..............eq(a), substitute x

y = 3/(1 +i*sqrt(2))
or
y = 3/(1 -i*sqrt(2))

solutions:
x = 1 +i*sqrt(2), y = 3/(1 +i*sqrt(2))
or
x = 1 - i*sqrt(2), y = 3/(1 -i*sqrt(2))

x^5 + y^5=(1 +i*sqrt(2))^5 + (3/(1 +i*sqrt(2)))^5=2
x^5 + y^5=(1 -i*sqrt(2))^5 + (3/(1 -i*sqrt(2)))^5=2

 

[nycmathguy]

Find x^5 + y^5.

x + y = 2...............eq1
xy = 3..............eq2

xy = 3..............eq2 solve for y

y = 3/x..............eq(a)

x + 3/x = 2...............eq1, solve for x
x^2/x + 3/x = 2
(x^2 + 3)/x = 2
x^2 + 3 = 2x

x^2 -2x+3=0........using quadratic formula you get

x = 1 +i*sqrt(2)
or
x = 1 - i*sqrt(2)

y = 3/x..............eq(a), substitute x

y = 3/(1 +i*sqrt(2))
or
y = 3/(1 -i*sqrt(2))

solutions:
x = 1 +i*sqrt(2), y = 3/(1 +i*sqrt(2))
or
x = 1 - i*sqrt(2), y = 3/(1 -i*sqrt(2))

x^5 + y^5=(1 +i*sqrt(2))^5 + (3/(1 +i*sqrt(2)))^5=2
x^5 + y^5=(1 -i*sqrt(2))^5 + (3/(1 -i*sqrt(2)))^5=2

Exactly my answer on the piece of paper I lost.