Question 1196483
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Given that {{{f(x)=5x^2-3x+7}}} and {{{f(g(x))=(5x^4/9)+(17x^2/3)+21}}}, find all possible values 
for the sum of the coefficients in the quadratic function g(x).
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            The key idea to solve this problem is to use the fact that the sum of coefficients 


            of any polynomial   p(x) = {{{a[n]*x^n + a[n-1]*x^(n-1) + ellipsis + a[1]*x + a[0]}}}   is the value 


            of this polynomial  p(1)  at   x = 1,  which is quite  OBVIOUS.



<pre>
Based on this idea, the sum of the coefficients in the quadratic function g(x) is g(1).


We don't know this value g(1), but from the problem we can calculate the composition  f(g(1))

by substituting x= 1 into the given formula for f(g(x).  We have then


    f(g(1)) = {{{(5*1^4)/9}}} + {{{(17*1^2)/3}}} + {{{21}}} = {{{5/9 + 17/3 + 21}}} = {{{5/9 + 51/9 + 21}}} = {{{56/9 + 21}}} = 27 2/9.


Now we can state that the value g(1), which we are looking for, is one of two possible roots of the equation

    f(x) = 27 2/9,  


or one of the two possible roots of the equation

    5x^2 - 3x + 7 = 27 2/9.


In standard quadratic form, this equation is

    5x^2 - 3x - 20 {{{2/9}}} = 0,


or, multiplying all the terms by 9, for convenience,

    45x^2 - 27x - 182 = 0.


To find the roots, use the quadratic formula.


The roots are  {{{210/90}}} = {{{7/3}}}  and  {{{-156/90}}} = {{{-26/15}}}.


So we conclude that the sum of the coefficients in the quadratic function g(x) is EITHER  {{{7/3}}} = 2{{{1/3}}}  OR  {{{-26/15}}} = -1{{{11/15}}}.    <U>ANSWER</U>
</pre>

Solved.


It is a nice Math circle level problem.