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Simultaneous equations

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Hi MHB,

I hope to gain some insights on how to solve this system of equations because I tried it many times to use trigonometric approach but to no avail...:mad:

Problem:

Solve the system of equations

$x^4-y^2-xy=4-\sqrt{15}$

$x^3+y^3-3x=5\sqrt{5}$

Attempt:

At first glance, $x^3-3x$ suggests the substitution of $x=\tan p$ so that $x^3-3x=\tan^3 p-3\tan p=(3\tan^2 p-1)(\tan 3p)$

and the second given equation can be rewritten as below, if we also let $y=\tan q$,

$x^3-3x=5\sqrt{5}-y^3$

$(3\tan^2 p-1)(\tan 3p)=5\sqrt{5}-\tan^3 q$

$\tan^3 q=5\sqrt{5}-(3\tan^2 p-1)(\tan 3p)$

and the first equation becomes

$x^4-y^2-xy=4-\sqrt{15}$

$\tan^4 p-\tan^2 q-\tan p \tan q=4-\sqrt{15}$

$\tan^2 q+\tan p \tan q+4-\sqrt{15}-\tan^4 p=0$

Even if I solve the equation above for $\tan q$ by the quadratic formula, I can see that this is of a futile attempt...

Could someone please help me to solve this very hard problem for me? Thanks in advance.:eek:
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
[graph]pqglddsffo[/graph] (Click on the diagram for an enlargement.)

The brown graph shows the first equation and the blue graph shows the second one. As they go off to infinity, the blue graph approximates the line $x+y=0$ and the brown graph approximates the parabolas $y = \pm x^2$. So I doubt whether there can be any solutions apart from the two crossing points in the diagram. The crossing point in the positive quadrant occurs when $x=\sqrt3,\ y=\sqrt5$ (obtained by guesswork!). I do not see how to locate the other crossing point.
 
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  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
[graph]pqglddsffo[/graph] (Click on the diagram for an enlargement.)

The brown graph shows the first equation and the blue graph shows the second one. As they go off to infinity, the blue graph approximates the line $x+y=0$ and the brown graph approximates the parabolas $y = \pm x^2$. So I doubt whether there can be any solutions apart from the two crossing points in the diagram. The crossing point in the positive quadrant occurs when $x=\sqrt3,\ y=\sqrt5$ (obtained by guesswork!). I do not see how to locate the other crossing point.
Thank you for this wonderful insight, Opalg! I will try to find the other intersection point and of course will add to this thread if I have found it.