All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Sum and Square Root Sum (Posted on 2010-09-25)
Determine all possible pair(s) (x,y) of real numbers that satisfy this set of equations:

x+y = 23, and:

√(x2 + 12*y) + √(y2 + 12*x) = 33

*** For an extra challenge, solve this puzzle without the aid of an online equation solver or a computer program.

 No Solution Yet Submitted by K Sengupta Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 No Subject Comment 3 of 3 |
The equations are symmetric in x and y, so I will try to use that to my advantage.

Start by squaring each side:
x^2 + 2xy + y^2  = 529
x^2 + 12*y + y^2 + 12*x + 2*sqrt[(xy)^2 + 12*y^3 + 12*x^3+ 144xy] = 1089

Rearrange and group into terms of (x^2+y^2), (x+y), and (xy):
(x^2+y^2) = 529-2xy
(x^2+y^2) + 12(x+y) + 2*sqrt[(xy)^2 + 12(x+y)(x^2+y^2-xy) + 144xy] = 1089

Now substitute (x^2+y^2) = 529-2xy and x+y=23 into the last equation to get an equation in terms of xy:
529-2xy + 12*23 + 2*sqrt[(xy)^2 + 12*23*(529-3xy) + 144*xy] = 1089

Isolate the sqrt expression and simplify:
2*sqrt[(xy)^2 + 146004 - 828xy + 144*xy] = 1089 - 529 + 2xy - 276
sqrt[(xy)^2 - 684*xy + 146004] = 142 + xy

Square each side and simplify:
(xy)^2 - 684*xy + 146004 = (xy)^2 + 284xy + 20164
968xy = 125840
xy = 130

Then the system xy=130, x+y=23 yields two solutions (x,y) = (10,13) and (13,10).

 Posted by Brian Smith on 2016-02-18 12:26:11

 Search: Search body:
Forums (0)