How to solve such a system of quadratic equations:x2+y2−xy=a2,x2+z2−xz=b2,y2+z2−yz=c2

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Kamora Campbell · Accepted Answer

Step 1(y+z−x)(y−z)=y2−z2−xy+xz=a2−b2(x+y−z)(x−y)=x2−y2−zx+zy=b2−c2(z+x−y)(z−x)=z2−x2−yz+yx=c2−a2Letu&amp;nbsp;=y−z,&amp;nbsp;v&amp;nbsp;=y+z,&amp;nbsp;α2&amp;nbsp;=a2−b2,&amp;nbsp;β2&amp;nbsp;=b2−c2,&amp;nbsp;γ2&amp;nbsp;=c2−a2, then(v−x)u=α2(x+u)(u+v−2x)=−2β2(x−u)(v−u−2x)=2γ2Eliminating&amp;nbsp;x=v−α2u, (unless&amp;nbsp;α=0=u)(uv+u2−α2)(u2+uv−2uv+2α2)=−2β2u2(uv−u2−α2)(uv−u2−2uv+2α2)=2γ2u2∴&amp;nbsp;(u2+uv−α2)(u2−uv+2α2)=−2β2u2(u2-uv+α2)(u2+uv-2α2)=2γ2u2∴&amp;nbsp;(u2+t12α2)2−(uv−t32α2)2=−2β2u2(u2−t12α2)2−(uv−t32α2)2=2γ2u2These two equations are actually the same:(u2−t12α2)2−(u2+t12α2)2=−2α2u2=2(β2+γ2)u2So we getv=(t32α2±(u2+t12α2)2+2β2u2)uIt seems we&#039;ve lost one degree of freedom. This is due to taking differences of the equations to start with. So reintroduce the y,z equation as(u+v)2+(u−v)2+(u+v)(u−v)=4c23u2+v2=4c2Substituting v in this equation gives a quartic in&amp;nbsp;u2, which can be solved.Hence find v, y, z, x in turn.Note: There may be 8 or 4 or 0 real solutions.

tabido8uvt · Accepted Answer

Step 1Solve the first and second equations as quadratic in y and z to get four pairs of solutionsy=12(x−4a2−3x2)z=12(x−4b2−3x2)y=12(x−4a2−3x2)z=12(x+4b2−3x2)y=12(x+4a2−3x2)z=12(x−4b2−3x2)y=12(x+4a2−3x2)z=12(x+4b2−3x2)Suppose that the last is the &quot;good&quot; one, replace in the third equation which is(a2+b2−c2)−5x24+x4(4a2−3x2+4b2−3x2)−144a2−3x24b2−3x2=0I suppose that multiple squaring steps would lead to an octic equation in x which, I hope, could be factorized.However, for given values of (a,b,c), I would prefer to graph the function to locate more or less the solution and after to use Newton method for the solution in x and then y and z.

How to solve such a system of quadratic

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