Question 1186259
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f(x) = x^3 - 2 and g(x) = x^2 - 5x. Solve gf(x) = 6.
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            I am  179%  sure that the problem in the post is written incorrectly.


            The correct writing is THIS


<pre>
              f(x) = x^3 - 2 and g(x) = x^2 - 5x,  solve gof(x) = 6.      (1)
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;So, &nbsp;the left side of the equation is the &nbsp;<U>composition</U> &nbsp;of polynomials &nbsp;gof, &nbsp;and not their product &nbsp;gf.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;After this correction, &nbsp;see my solution below.



<pre>
Introduce new variable y = x^3 - 2.


Then the given equation (1)  takes the form


    y^2 - 5y = 6

or

    y^2 - 5y - 6 = 0.


Factor left side

    (y-6)*(y+1) = 0,


which gives the roots  y= 6 and y= -1.



If  y= 6,   then  x^3 - 2 = 6,   x^3 = 6 + 2 = 8,   which implies   x = {{{root(3,8)}}} = 2.


If  y= -1,  then  x^3 - 2 = -1,  x^3 = -1 + 2 = 1,  which implies   x = {{{root(3,1)}}} = 1.



So, the real roots of the equation (1) are the values 1 and/or 2.



If you want to get all complex roots of equation (1), you should obtain complex roots of equations

    x^3 = 8  and  x^3 = 1.


They are  x = {{{2*cis(2pi/3)}}} and  {{{2*cis(4pi/3)}}}  for equation x^3 = 8,  and  x = {{{cis(2pi/3)}}} and  {{{cis(4pi/3)}}}  for equation x^3 = 1.


Thus the full list of the solutions to equation (1) is


     1,  {{{cis(2pi/3)}}},  {{{cis(4pi/3)}}},  2,  {{{2*cis(2pi/3)}}}  and  {{{2*cis(4pi/3)}}}.
</pre>

Solved.