# Sum of two numbers

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
Homework Helper
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.

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

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

1/x+ 1/y= 900.
i do not think this part is correct!!

symbolipoint
Homework Helper
Gold Member
Yes, imaginary solutions. Starting from the form, $$$x^2 - 50x + 900 = 0$$$, and directly using general quadratic equation solution, part of the expression will contain $$$\sqrt {2500 - 3600}$$$, which is imaginary.

Last edited:
Yes, imaginary solutions. Starting from the form, $$$x^2 - 50x + 900 = 0$$$, and directly using general quadratic equation solution, part of the expression will contain $$$\sqrt {2500 - 3600}$$$, 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
Homework Helper
Gold Member
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
Homework Helper
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
Homework Helper
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