The number of roots of the system equation: x+y=1, x^5+y^5=11?

Question

Apart from the method of substituting y=1-x into x^5+y^5=11, then expand the binomial and factor the 4th degree polynomial to solve for the root, is there any other faster way to find the answer?

? · Answer

Two real roots. I used fooplot.com to find them. The curve x^5 + y^5 = 11 is asymptotic to y+x=0, but the curve has a "bulge" that protrudes into the 1st quadrant in the vicinity of x = (11/2)^(1/5), y = (11/2)^(1/5). Since (11/2)^(1/5) is about 1.406, obviously x+y > 1 at the point just mentioned. The curve "falls" back towards the diagonal x+y = 0 as you go into the 2nd and 4th quadrant. The roots of your system are somewhat near (1.6,-0.6) and (-0.6,1.6).

The short answer to your final question is: graphically! Look for the intersection points of the line with the curve.

The short answer to your final question is: graphically!  Look for the intersection points of the line with the curve.

The number of roots of the system equation: x+y=1, x^5+y^5=11?

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