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Quad composition (Posted on 2019-03-16) Difficulty: 3 of 5
Let f(x)=x2+10x+20, find the number of real solutions satisfying f(f(f(f(x))))=0

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution calculator solution | Comment 1 of 4
What must f(f(f(x))) be in order to get the next level to be zero?

(10 +/- sqrt(100-20))/2 

5 +/- sqrt(20)

Then set x^2 + 10x + 20 equal to the above to find the values for f(f(x)):

x^2 + 10x + 15 - sqrt(20)      or     x^2 + 10x + 15 + sqrt(20)

The algebra would seem to get overly complicated.

I'll switch to a graphing calculator.

Setting the TI-84 Plus CE: Y1= x^2 + 10x + 20, and Y2 = Y1(Y1(Y1(Y1))), graphing and tracing and using the calc function zero, finds, first a squared off "parabola" centered on x=-5 with minimum y=-5. The zero calc function finds zeros at x=-6.105823 and x=-3.894177. These two values are equidistant around x=-5. Asking for values outside the range of these x values result in ever increasing y values requiring scientific notation with higher and higher values of 10^k.

There are two real solutions.


  Posted by Charlie on 2019-03-16 13:20:33
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