Find Value of x+y Reciprocals: 50 & 900

In summary, the sum of two numbers is 50, and their product is 900. The exact value of the sum of their reciprocals is 5/90.
  • #1
rocomath
1,755
1
The sume of two numbers is 50, and their product is 900. Find the exact value of the sum of their reciprocals.

[tex]x+y=50[/tex]

[tex]xy=900[/tex]

[tex]x^2-50x+900[/tex]

Imaginary solutions?
 
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  • #2
Blast! I just spent 10 minutes writing out an explanation of why "if x+y= 50, xy cannot be more than 225" when I suddenly noticed the word "reciprocals"!

x+ y= 50 and 1/x+ 1/y= 900.
 
  • #3
HallsofIvy said:
Blast! I just spent 10 minutes writing out an explanation of why "if x+y= 50, xy cannot be more than 225" when I suddenly noticed the .
did u mean here xy cannot be more than 625, because let x=y=25, then
xy=25*25=625
 
  • #4
to the op
like halls started
1/x+1/y=(x+y)/xy =50/900
 
  • #5
HallsofIvy said:
1/x+ 1/y= 900.
i do not think this part is correct!
 
  • #6
Yes, imaginary solutions. Starting from the form, [tex] \[
x^2 - 50x + 900 = 0
\]
[/tex], and directly using general quadratic equation solution, part of the expression will contain [tex] \[
\sqrt {2500 - 3600}
\]
[/tex], which is imaginary.
 
Last edited:
  • #7
symbolipoint said:
Yes, imaginary solutions. Starting from the form, [tex] \[
x^2 - 50x + 900 = 0
\]
[/tex], and directly using general quadratic equation solution, part of the expression will contain [tex] \[
\sqrt {2500 - 3600}
\]
[/tex], which is imaginary.

well then, can't one do what i did?

let 1/x be the reciprocal for x, and 1/y the reciprocal of y, so we want to find the sum of them, and we have:

1/x+1/y=(x+y)/xy=50/900=5/90 , what is wrong with this?
 
  • #8
Quote:
Originally Posted by symbolipoint
Yes, imaginary solutions. Starting from the form, , and directly using general quadratic equation solution, part of the expression will contain , which is imaginary.

well then, can't one do what i did?

let 1/x be the reciprocal for x, and 1/y the reciprocal of y, so we want to find the sum of them, and we have:

1/x+1/y=(x+y)/xy=50/900=5/90 , what is wrong with this?

Not sure, since I did not carry the solution far enough. I only developed a solution for x and y. Try taking those solutions and then yourself carry farther to the extent of the question asked (about the sum of the reciprocals) and find out how it works. I did not disagree with you; I merely did not take the intermediate solutions far enough for the original question.
 
  • #9
sutupidmath said:
well then, can't one do what i did?

let 1/x be the reciprocal for x, and 1/y the reciprocal of y, so we want to find the sum of them, and we have:

1/x+1/y=(x+y)/xy=50/900=5/90 , what is wrong with this?

That's the clever way to do it. If you find the roots like, x=25+5*sqrt(11)*i and y=conjugate(x), invert them and add them, you'll get... 50/900.
 
  • #10
HallsofIvy said:
Blast! I just spent 10 minutes writing out an explanation of why "if x+y= 50, xy cannot be more than 225" when I suddenly noticed the word "reciprocals"!

x+ y= 50 and 1/x+ 1/y= 900.

sutupidmath said:
did u mean here xy cannot be more than 625, because let x=y=25, then
xy=25*25=625
Well, I did a bit more than just calculate that value. Assuming, as was implied by the work but is wrong, that the equations are x+ y= 50 and xy= 900, replace y with 50-x so x(50- x)= 900. The left side is a quadratic function. Completing the square, I found that the vertex of its graph is at (25, 625) and so the value on the left side cannot be more than 625.
 
  • #11
sutupidmath said:
i do not think this part is correct!
Right. I got so annoyed at doing useless work I just wrote it out wrong. Your solution is very nice.
 
  • #12
HallsofIvy said:
Well, I did a bit more than just calculate that value. Assuming, as was implied by the work but is wrong, that the equations are x+ y= 50 and xy= 900, replace y with 50-x so x(50- x)= 900. The left side is a quadratic function. Completing the square, I found that the vertex of its graph is at (25, 625) and so the value on the left side cannot be more than 625.
Yeah, but you wrote there "cannot be more than 225", and i know this was a typo, so i just wanted to point it out!
 
  • #13
Wow, very nice! Thanks :-]
 

Related to Find Value of x+y Reciprocals: 50 & 900

1. What is the formula for finding the value of x+y reciprocals?

The formula for finding the value of x+y reciprocals is 1/(x+y).

2. How do I solve for x and y in the equation 1/(x+y) = 50 & 900?

To solve for x and y in this equation, you would first need to set up two equations using the given information. 1/x = 50 and 1/y = 900. Then, you can solve for x and y individually using basic algebraic manipulation.

3. Can I use any numbers for x and y in the equation 1/(x+y) = 50 & 900?

Yes, you can use any numbers for x and y in this equation. However, for the equation to be true, x and y must be reciprocals of each other. For example, x=2 and y=1/2 would work, as 1/2 + 2 = 2.5 and 1/2 = 0.5.

4. What is the significance of finding the value of x+y reciprocals?

Finding the value of x+y reciprocals can be useful in solving mathematical equations and understanding the relationship between two numbers. It can also be used to solve real-world problems, such as finding the total resistance in an electrical circuit.

5. How can I apply the concept of x+y reciprocals in other scientific fields?

The concept of x+y reciprocals can be applied in various scientific fields, such as physics, chemistry, and biology. For example, in physics, it can be used to calculate the total energy in a system. In chemistry, it can be used to determine the molar concentration of a solution. In biology, it can be used to calculate the total population size of a species.

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