Solving 4th Degree Equation: √x+y=7, √y+x=11

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Discussion Overview

The discussion revolves around solving a system of equations involving square roots: √x + y = 7 and √y + x = 11. Participants explore methods for solving these equations, which lead to a fourth-degree polynomial or two quadratic equations. The conversation includes various approaches and insights into the complexity of the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes that using the quadratic formula for a fourth-degree equation is impossible and suggests the equations can be transformed into two quadratic equations.
  • Another participant proposes isolating y in one equation and substituting it into the other, leading to a quartic equation that may be complicated to solve.
  • A third participant references a computational tool that indicates one real solution (x=9) and one complex solution, without detailing the complex solution.
  • Further discussion confirms x=9 as a solution but raises questions about the existence of additional solutions, including potential extraneous solutions from the quartic equation.
  • One participant shares a method of substituting variables (x=l², y=n²) to eliminate square roots, resulting in two quadratic equations, ultimately finding x=9 and y=4.

Areas of Agreement / Disagreement

Participants generally agree that the problem leads to a quartic equation and that x=9 is a solution. However, there is uncertainty regarding the existence of additional solutions, with some participants questioning whether there are more complex solutions or if x=9 is the only solution.

Contextual Notes

Some participants express uncertainty about the nature of the solutions, including the possibility of extraneous solutions arising from squaring the equations. The discussion also highlights the complexity of solving quartic equations and the potential for rational roots.

Who May Find This Useful

Readers interested in algebra, particularly those exploring methods for solving polynomial equations and the implications of square roots in equations, may find this discussion relevant.

obing
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√x+y=7
√y+x=11
solving this equation using x=-b√b2-4ac/2a for finding the root is impossible

the equation turns out to be of 4th degree

or two quadratic eq. can anybody help

this isn't a homework , just wanted to know ...;)
 
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If you're trying to solve for x in these equations, isolate y in either the first or second and then substitute it into the other equation.

[itex]\sqrt{x}+y=7[/itex] (1)

[itex]\sqrt{y}+x=11[/itex] (2)

Rearranging (2) to make y the subject:

[itex]\sqrt{y}=11-x[/itex]

[itex]y=(11-x)^2[/itex]

Now substitute into (1):

[itex]\sqrt{x}+(11-x)^2=7[/itex]

To simplify this it's important to try get rid of the square root. This would imply squaring, but remember if we just squared both sides as it is, something like [itex](\sqrt{a}+b)^2=a+2b\sqrt{a}+b^2[/itex] so we don't really get rid of the root, we just create more problems.

So instead, isolate the [itex]\sqrt{x}[/itex] and then square both sides.

Yes you will get a quartic (4th degree) equation and it is generally very complicated to solve, unless you're in luck and this question was devised to have nice, rational roots in which case you could then use the rational root theorem to find them.
But I doubt this since it sounds like you created the problem yourself.
 
http://www.wolframalpha.com/input/?i=sqrt%28x%29%3D7-%2811-x%29^2

According to this, one real solution, 9, and one complex solution, which I won't attempt to copy...
 
Certainly, x=9 is a solution.

If you continue to solve and finally end with the quartic and then solve that for x, you're given 4 real solutions. Some of these however could be extraneous solutions.

But look at what wolfram gives as solutions to the original system of equations:
http://www.wolframalpha.com/input/?i=sqrt(x)+y=7,+sqrt(y)+x=11"

Also, graphmatica draws [itex]\sqrt{y}+x=11[/itex] as the parabola [itex]y=(11-x)^2[/itex] which is fine except that for x>11 it does not satisfy the first equation. And this really stumped me at first.

Well, I'm stumped. Is the only solution x=9 or are there more complex solutions?
 
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thanks guys :)

ive found a way out


if we take x= l2

and y=n2

then √l2 +y= 7 (√x+y=7)


√n2+x =11 (√y+x=11)

which becomes l2+y-7=0 (getting rid of root)



n2+x-11=0

2 quadratic equn. :)

ans sloving it gives x=9 y=4
 

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