Solve simple looking system of equations

[Mondo]

Hey,

I stumbled upon simply looking but no to simple solvable system of equations [math]\begin{cases} y^2 + 1 = (4-x)^2 \\ x = (4-x)/y \end{cases}[/math]I get to the point [math]x^4 - 8x^3+14x^2 + 8x - 16 = 0[/math][math]x^2(x^2 -8x+14) + 8x - 16 = 0[/math]But from this point I have no idea how to proceed to obtain all possible solutions.
Thanks for help!

 

[Deleted member 4993]

Hey,

I stumbled upon simply looking but no to simple solvable system of equations [math]\begin{cases} y^2 + 1 = (4-x)^2 \\ x = (4-x)/y \end{cases}[/math]I get to the point [math]x^4 - 8x^3+14x^2 + 8x - 16 = 0[/math][math]x^2(x^2 -8x+14) + 8x - 16 = 0[/math]But from this point I have no idea how to proceed to obtain all possible solutions.
Thanks for help!

I would first substitute:

u = 4 - x and get

y^2 + 1 = u^2 .... and

y = u/(4 - u)

Continue .....

 

[Dr.Peterson]

Hey,

I stumbled upon simply looking but no to simple solvable system of equations [math]\begin{cases} y^2 + 1 = (4-x)^2 \\ x = (4-x)/y \end{cases}[/math]I get to the point [math]x^4 - 8x^3+14x^2 + 8x - 16 = 0[/math][math]x^2(x^2 -8x+14) + 8x - 16 = 0[/math]But from this point I have no idea how to proceed to obtain all possible solutions.
Thanks for help!

Interestingly, when I follow SK's good suggestion, I get the same quartic equation! That implies symmetry about x = 2. So the next thing to try is the substitution z = 2 - x; this produces a quartic that can be easily solved.

Fascinating problem! Where did it come from?

 

[Mondo]

Interestingly, when I follow SK's good suggestion, I get the same quartic equation! That implies symmetry about x = 2.

Hmm how do you see this symmetry about x = 2?

So the next thing to try is the substitution z = 2 - x; this produces a quartic that can be easily solved.

I did like this [math]z = 2-x \\ y^2 + 1 = (2+z)^2 \\ y = \frac{2+z}{2-z}[/math] which leads me to [math]u^2(u^2-10)+8 = 0[/math] which again I have trouble to solve.

Fascinating problem! Where did it come from?

heh, the origin is nothing special - I just wrote two equations describing simple geometric relationship between two similar triangles to find unknown side.

 

[Dr.Peterson]

Hmm how do you see this symmetry about x = 2?


I did like this [math]z = 2-x \\ y^2 + 1 = (2+z)^2 \\ y = \frac{2+z}{2-z}[/math] which leads me to [math]u^2(u^2-10)+8 = 0[/math] which again I have trouble to solve.

Now do another substitution, if you need to: let [imath]v = u^2[/imath]. Then expand and refactor. (The form you have it in isn't helpful.)

I'll tell you a secret: I didn't immediately see the symmetry. I just got the same quartic equation, figured I'd done something wrong, and moved on to something else. Then later I tried graphing the equations and saw visually that the quartic looked symmetrical, and only then realized that fact had been visible in the equation. You see, the substitution u = 4 - x is a reflection about the line x = 2, because it moves x=0 to x=4 and vice versa. So if it produces an identical equation, then the graph is symmetrical.

The reason for doing the new substitution is to turn the equation into an even function, with only even powers of the variable. Then we know it can be factored in powers of the square, i.e. [imath]v = u^2[/imath].

heh, the origin is nothing special - I just wrote two equations describing simple geometric relationship between two similar triangles to find unknown side.

I'd like to see the original geometric problem; it probably has some features as interesting as the equations, and perhaps could be solved more easily by a different method.

 

[Deleted member 4993]

Hmm how do you see this symmetry about x = 2?


I did like this [math]z = 2-x \\ y^2 + 1 = (2+z)^2 \\ y = \frac{2+z}{2-z}[/math] which leads me to [math]u^2(u^2-10)+8 = 0[/math] which again I have trouble to solve.

heh, the origin is nothing special - I just wrote two equations describing simple geometric relationship between two similar triangles to find unknown side.

u^2(u^2-10)+8 = 0

Assuming your work is correct upto here - substitute v = u²

v² - 10v + 8= 0 (quadratic in 'v')

v = 5 ± √(17) ...... continue.....

 

[Mondo]

Then later I tried graphing the equations and saw visually that the quartic looked symmetrical

What tool do you use to obtain the graphical representation of an equation?

The reason for doing the new substitution is to turn the equation into an even function, with only even powers of the variable. Then we know it can be factored in powers of the square, i.e. [math]v = u^2v=u2.[/math]

Hmm I am not sure if got it. Yes I was able to solve the equation using your suggestion of substitution [math]y = 2-x \\ y ^2 + 1 = (2+u)^2[/math] but can you elaborate how did you came up with the idea of substitution [imath]u = 2-x[/imath]?

I'd like to see the original geometric problem; it probably has some features as interesting as the equations, and perhaps could be solved more easily by a different method.

Sure, here it is:


l = 4, c = 1, y = ? (yes the figure inside is a square of side c)
So my solution relies on Pythagoras formula + simple relation between sides of two similar triangles. Would you solve it differently?

Subhotosh Khan, thanks I managed to solve that.

 

[Dr.Peterson]

What tool do you use to obtain the graphical representation of an equation?

In this case I used Desmos.

but can you elaborate how did you came up with the idea of substitution [imath]u = 2-x[/imath]?

Now that I think about it, it would have made more sense to suggest shifting the graph left 2 units by replacing x with u+2 so that u = x-2 rather than 2-x, which involves a reflection along with the shift; but either substitution results in this:



Green is the original quartic, which is symmetrical about x=2, and red is the new one, which is symmetrical about the y-axis. I think I let my thinking be affected by the suggestion of 4-x.

Sure, here it is:
View attachment 28623

l = 4, c = 1, y = ? (yes the figure inside is a square of side c)
So my solution relies on Pythagoras formula + simple relation between sides of two similar triangles. Would you solve it differently?

I would do something similar. My first thought, using just the variable y you show, is to write the proportion [imath]\frac{y+1}{4}=\frac{1}{\sqrt{y^2+1}}[/imath], which leads directly to the quartic [imath](y+1)^2(y^2+1)=16y^2[/imath]; that doesn't turn out to be symmetrical.

Looking at your equations, it appears that you took x to be the lower part of the hypotenuse l; if I used just that x, I'd have [imath]\frac{4-x}{1}=\frac{x}{\sqrt{x^2-1}}[/imath], which leads directly to the quartic [imath](4-x)^2(x^2-1)=x^2[/imath], which is yours.

So I'm just skipping the two-variable step. The symmetry of the problem makes it clear that there will be at least two solutions, because putting x for the upper part of the hypotenuse leads to the same equation. This also suggests (though I don't promise I'd notice it) that the solutions would be symmetrical about x=2. (That's not true of y.)

I know problems that look very similar that result in quadric equations, so I don't expect that there is a nice way to avoid that.

 

[Mondo]

Thanks for desmos recomendation - nice tool. Funny we can read out the solution right from the graph. Only one out of 5 solutions makes sense in regards to the problem these equations describe

 

[Dr.Peterson]

Thanks for desmos recomendation - nice tool. Funny we can read out the solution right from the graph. Only one out of 5 solutions makes sense in regards to the problem these equations describe


View attachment 28631

What five solutions do you see? There are only four! And of those, two are in the first quadrant, and correspond to two solutions that are related in an obvious way:





Also, of course, Desmos can't give exact answers; for that, I used Wolfram Alpha (in addition to solving the way we've discussed).

 

[Mondo]

Yep, my mistake - there are only four solutions. Thanks, you explained it very well!