- #1

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

[tex]xy=900[/tex]

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

Imaginary solutions?

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- Thread starter rocomath
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- #1

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

[tex]xy=900[/tex]

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

Imaginary solutions?

- #2

HallsofIvy

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

- #3

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did u mean here xy cannot be more than 625, because let x=y=25, thenBlast!! 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 .

xy=25*25=625

- #4

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to the op

like halls started

1/x+1/y=(x+y)/xy =50/900

like halls started

1/x+1/y=(x+y)/xy =50/900

- #5

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

- #6

symbolipoint

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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.

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

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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

symbolipoint

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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

Dick

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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

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

Well, I did adid u mean here xy cannot be more than 625, because let x=y=25, then

xy=25*25=625

- #11

HallsofIvy

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Right. I got so annoyed at doing useless work I just wrote it out wrong. Your solution is very nice.i do not think this part is correct!!

- #12

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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!Well, I did abitmore 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.

- #13

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Wow, very nice! Thanks :-]

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