Find Value of x+y Reciprocals: 50 & 900

[rocomath]

The sume of two numbers is 50, and their product is 900. Find the exact value of the sum of their reciprocals.

$$x+y=50$$

$$xy=900$$

$$x^2-50x+900$$

Imaginary solutions?

 

[HallsofIvy]

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]

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

 

[sutupidmath]

to the op
like halls started
1/x+1/y=(x+y)/xy =50/900

 

[sutupidmath]

1/x+ 1/y= 900.

i do not think this part is correct!

 

[symbolipoint]

Yes, imaginary solutions. Starting from the form, $$\[<br /> x^2 - 50x + 900 = 0<br /> \]$$, and directly using general quadratic equation solution, part of the expression will contain $$\[<br /> \sqrt {2500 - 3600} <br /> \]$$, which is imaginary.

 

[sutupidmath]

Yes, imaginary solutions. Starting from the form, $$\[<br /> x^2 - 50x + 900 = 0<br /> \]$$, and directly using general quadratic equation solution, part of the expression will contain $$\[<br /> \sqrt {2500 - 3600} <br /> \]$$, 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?

 

[symbolipoint]

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.

 

[Dick]

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.

 

[HallsofIvy]

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.

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.

 

[HallsofIvy]

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.

 

[sutupidmath]

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!

 

[rocomath]

Wow, very nice! Thanks :-]