+934 I thinks the answer is 1 because a can equal 5 and -4. I didn't realize that a could be negative.
Hope it helps!
+26367 Let \(f(x)=3x+2\) and \(g(x)=ax+b\), for some constants \(a\) and b.
If \(ab=20\) and \(f(g(x))=g(f(x))\) for \(x=0,1,2,\ldots ,9\), find the sum of all possible values of \(a\).
\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{3x+2} \quad &| \quad x=g(x) \\\\ f(g(x)) &=& 3g(x)+2 \quad &| \quad g(x)=ax+b \\ f(g(x)) &=& 3(ax+b)+2 \\ f(g(x)) &=& 3ax+3b+2 \quad &| \quad b=\dfrac{20}{a} \\ f(g(x)) &=& 3ax+3*\dfrac{20}{a}+2 \\ \mathbf{f(g(x))} &=& \mathbf{3ax+ \dfrac{60}{a}+2} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{g(x)} &=& \mathbf{ax+b} \quad &| \quad x=f(x) \\\\ g(f(x)) &=& af(x)+b \quad &| \quad f(x)=3x+2 \\ g(f(x)) &=& a(3x+2)+b \\ g(f(x)) &=& 3ax+2a+b \quad &| \quad b=\dfrac{20}{a} \\ \mathbf{g(f(x))} &=& \mathbf{3ax+2a+\dfrac{20}{a}} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \mathbf{ f(g(x)) } &=& \mathbf{g(f(x))} \\\\ 3ax+ \dfrac{60}{a}+2 &=& 3ax+2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}+2 &=& 2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}-\dfrac{20}{a}+2 &=& 2a \\\\ \dfrac{40}{a} +2 &=& 2a \quad &| \quad * a \\\\ 40 +2a &=& 2a^2 \quad &| \quad : 2 \\ 20 + a &=& a^2 \\ \mathbf{ a^2-a-20} &=& \mathbf{0} \\\\ a _{1,2} &=& \dfrac{1 \pm \sqrt{1-4(-20)} }{2} \\ a _{1,2} &=& \dfrac{1 \pm \sqrt{81} }{2} \\\\ a _{1,2} &=& \dfrac{1 \pm 9 }{2} \\\\ a_1 = \dfrac{1}{2} + \dfrac{9}{2} && a_2 = \dfrac{1}{2} - \dfrac{9}{2} \\ a_1+a_2 &=& 2\times \dfrac{1}{2} \\ \mathbf{a_1+a_2} &=& \mathbf{1} \\ \hline \end{array}\)
