MEI Maths Item of the Month
February 2015
1, 2, 3, 4
Find two quadratic functions f(x), g(x) so the equation f(g(x)) = 0 has the four roots x = 1, 2, 3,
4.
Is it possible to find three quadratic functions f(x), g(x), h(x) so the equation f(g(h(x))) = 0 has
the eight roots x = 1, 2, 3, 4, 5, 6, 7, 8?
Solution
f( x)  x( x  2) and g( x)  ( x  2)( x  3) gives f(g( x))  x 4  10 x 3  35 x 2  50 x  24 which
has roots x = 1, 2, 3, 4.
It is possible to find a function f( x) for any g( x) that is symmetrical about x = 2.5. For
example if g( x)  2 x 2  10 x  9 then the horizontal line through the curve when x = 1 and
x = 2 will also cut the curve when x = 3 and x = 4.
Setting f( x)  ( x  1)( x  3) gives f(g( x))  4 x 4  40 x 3  140 x 2  200 x  96 which has roots
x = 1, 2, 3, 4.
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MEI Maths Item of the Month
To find three quadratic functions f(x), g(x), h(x) so the equation f(g(h(x))) = 0 has the eight
roots x = 1, 2, 3, 4, 5, 6, 7, 8 it would be necessary to find a quartic p(x) such that one
horizontal line cuts the curve at x = 1, 4, 5, 8 and a second cuts the curve at x = 2, 3, 6, 7.
2


9
p(x) would need to be symmetrical about x = 4.5, i.e. p( x)  a   x    q   r


2


Solving p(1) = p(4) for q gives q  
25
.
4
2

9  25 
p( x)  a   x      r

2
4 

 ax 4  18ax3  109ax 2  252ax  196a  r
p'( x)  4ax3  54ax 2  218ax  252a
p'(x) = 0 when x = 2, 4.5 and 7. Therefore p(2) ≠ p(3) and hence it is not possible to find such
a quartic so it is not possible to find three quadratic functions f(x), g(x), h(x) so the equation
f(g(h(x))) = 0 has the eight roots x = 1, 2, 3, 4, 5, 6, 7, 8.
An example of a quartic that is symmetrical about x = 4.5, and has p(1) = p(4) is shown below
with the turning points at x = 2 and 7 marked.
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