+0

Help!

+3
231
3
+339

https://web2.0calc.com/questions/help-pls_111#r2

Apr 7, 2020

#1
+924
+4

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!

Apr 7, 2020
edited by HELPMEEEEEEEEEEEEE  Apr 7, 2020
#2
+25644
+4

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 $$a$$.

$$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{3x+2} \quad &| \quad x=g(x) \\\\ f(g(x)) &=& 3g(x)+2 \quad &| \quad g(x)=ax+b \\ f(g(x)) &=& 3(ax+b)+2 \\ f(g(x)) &=& 3ax+3b+2 \quad &| \quad b=\dfrac{20}{a} \\ f(g(x)) &=& 3ax+3*\dfrac{20}{a}+2 \\ \mathbf{f(g(x))} &=& \mathbf{3ax+ \dfrac{60}{a}+2} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{g(x)} &=& \mathbf{ax+b} \quad &| \quad x=f(x) \\\\ g(f(x)) &=& af(x)+b \quad &| \quad f(x)=3x+2 \\ g(f(x)) &=& a(3x+2)+b \\ g(f(x)) &=& 3ax+2a+b \quad &| \quad b=\dfrac{20}{a} \\ \mathbf{g(f(x))} &=& \mathbf{3ax+2a+\dfrac{20}{a}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{ f(g(x)) } &=& \mathbf{g(f(x))} \\\\ 3ax+ \dfrac{60}{a}+2 &=& 3ax+2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}+2 &=& 2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}-\dfrac{20}{a}+2 &=& 2a \\\\ \dfrac{40}{a} +2 &=& 2a \quad &| \quad * a \\\\ 40 +2a &=& 2a^2 \quad &| \quad : 2 \\ 20 + a &=& a^2 \\ \mathbf{ a^2-a-20} &=& \mathbf{0} \\\\ a _{1,2} &=& \dfrac{1 \pm \sqrt{1-4(-20)} }{2} \\ a _{1,2} &=& \dfrac{1 \pm \sqrt{81} }{2} \\\\ a _{1,2} &=& \dfrac{1 \pm 9 }{2} \\\\ a_1 = \dfrac{1}{2} + \dfrac{9}{2} && a_2 = \dfrac{1}{2} - \dfrac{9}{2} \\ a_1+a_2 &=& 2\times \dfrac{1}{2} \\ \mathbf{a_1+a_2} &=& \mathbf{1} \\ \hline \end{array}$$

Apr 8, 2020
edited by heureka  Apr 8, 2020
#3
+339
+4

Thank you!

Apr 8, 2020