This question already has an answer here:

A friend of Mine gave me a system of two equations and asked me to solve them $\rightarrow$

$$\sqrt{x}+y=11~~ ...1$$ $$\sqrt{y}+x=7~~ ...2$$

I tried to solve them manually and got this horrendously complicated fourth degree equation $\rightarrow$

$$\begin{align*} y &= (7-x)^2 ~...\mbox{(from 2)} \\ y &= 49 - 14 x + x^2 \\ \implies 11&= \sqrt{x}+ 49 - 14 x + x^2 ...(\mbox{from 1)}\\ \implies~~ 0&=x^4-28x^3+272x^2-1065x+1444 \end{align*}$$

Solving this wasn't exactly my piece of cake but I could tell that one of Solutions would have been 9 and 4

But my friend kept asking for a formal solution.

I tried plotting the equations and here's what I got $\rightarrow$

enter image description here

So the equations had two pairs of solutions (real ones).

Maybe, Just maybe I think these could be solved using approximations.

So How do i solve them using a formal method (Calculus,Algebra,Real Analysis...)

P.S. I'm In high-school.


marked as duplicate by Misha Lavrov, Parcly Taxel, mlc, Rohan, Namaste algebra-precalculus Dec 6 '17 at 15:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ From the fourth degree equation, you can use Ferrari's method (en.wikipedia.org/wiki/Quartic_function) to solve it. $\endgroup$ – guaraqe Jun 1 '12 at 15:05
  • $\begingroup$ In this question a solution for the same system (with $x,y$ interchanged) is asked. As such the present question is a duplicate. $$\sqrt{x} + y = 7$$ $$\sqrt{y} + x =11.$$ $\endgroup$ – Américo Tavares Jun 1 '12 at 15:41
  • 2
    $\begingroup$ @The-Ever-Kid: OP stands for Original Post(er). $\endgroup$ – Américo Tavares Jun 1 '12 at 15:49
  • 2
    $\begingroup$ @Thomas: My answer to the second question I've linked to in my second comment above gives an algebraic solution. $\endgroup$ – Américo Tavares Jun 1 '12 at 17:03
  • 1
    $\begingroup$ @the-ever-kid sorry, I got confused with the names. I'm not sure what you're getting at with your comment about gnuplot. Just because yours is from mathematica doesn't make it more right (mine was from Maple btw). Just to prove it to you though, here's the same implicit graph from the original system of equations from Mathematica: imgur.com/dPqq1 which you can draw yourself with: $$ContourPlot[{y + (x)^{(1/2)} == 11 , x + (y)^{(1/2)} == 7}, {x, 0, 10}, {y, 0, 10}]$$ also, wolframalpha: wolframalpha.com/input/… $\endgroup$ – Andrew Christianson Jun 1 '12 at 18:56

Assume $x$ and $y$ are integers. Notice that, in this case, if $\sqrt x +y=11$, an integer, then $\sqrt x $ must be an integer. A similar argument can be made for $y$. So if they're integers then they're both perfect squares. Rephrasing in terms of the square roots (still integers) $X=\sqrt x,Y=\sqrt y$ $$X+Y^2=11$$ $$Y+X^2=7$$ subtracting the second equation from the first: $$X-Y+Y^2-X^2=4$$ $$(X-Y)+(Y-X)(Y+X)=4$$ $$(Y-X)(X+Y-1)=4$$ Both of the brackets are integers, so the only values they can take are the factors of $4$. So either $$Y-X=2,X+Y-1=2$$ or $$Y-X=4,X+Y-1=1$$ or$$Y-X=1,X+Y-1=4$$ Solving each of these is simple. The only one that gives positive integer values (the conditions of our little set up here) is the $3^{rd}$ one, which gives the answer you found. Keep in mind that there's nothing wrong with guessing and playing around with the problem first, then coming to a more structured argument later. If you want a full analytic solution you could use the quartic equation on the one you have and rule out the other solutions as involving the wrong branches of $\sqrt x$, but it's pointlessly messy.

  • $\begingroup$ How come the graphs intersect twice and the points other than 4,9 are not integers..... $\endgroup$ – The-Ever-Kid Jun 1 '12 at 15:35
  • $\begingroup$ Mathematica plots figured two solutions other than 4,9 -> 9.8051180869527 and 7.8686874817494 $\endgroup$ – The-Ever-Kid Jun 1 '12 at 15:41
  • $\begingroup$ it's the bit about branches of $\sqrt x$ i mentioned at the end. $\sqrt x$ is defined as the positive number whose square is $x$. So $\sqrt 4=x$, gives $x=2$. But $x^2=4$ gives us two solutions, $\pm 2$. The extra solution isn't part of our original problem and not something we care about. Rephrasing your question in terms of squares and not square roots introduces extra 'solutions' that don't solve your original problem. The extra answer mathematica produced only works if you take the negative square root for the second equation, which isn't how the square root function is defined. $\endgroup$ – Robert Mastragostino Jun 1 '12 at 16:16
  • $\begingroup$ look at the image that is a perfectly real and positive graph $\endgroup$ – The-Ever-Kid Jun 1 '12 at 16:35
  • $\begingroup$ the solutions arnt negative $\endgroup$ – The-Ever-Kid Jun 1 '12 at 16:35

Once you guessed the solutions, you can easily prove that there are no others. Rewrite the equations as $y=11-\sqrt x=F(x)$ and $x=7-\sqrt y=G(y)$. Note that both $x,y\le 11$, so their square roots are at most $4$, which means that $x,y\ge 3$. Now just observe that $z\mapsto \sqrt z$ is a contraction on $[3,\infty)$ (the difference of values is less than the difference of arguments). Thus, $F$ and $G$ are also contractions whence if we had two different solutions $(x_1,y_1)$ and $(x_2,y_2)$, we would get $$ |x_1-x_2|=|G(y_1)-G(y_2)|<|y_1-y_2|=|F(x_1)-F(x_2)|<|x_1-x_2| $$ which is absurd.


Not the answer you're looking for? Browse other questions tagged or ask your own question.