Question #0372d

1 Answer
Mar 13, 2016

Answer:

#a=3#
#f[g(2)]=4#
imaginary roots#=(-1+-sqrt(3)i)/2#

Explanation:

Part A
#1#. Given the following, substitute #color(orange)(f(x))# into #color(blue)x# in #color(blue)(g(x))#.

#color(orange)(f(x)=ax+1)#

#color(blue)(g(x)=x^2-3)#

#color(blue)(g[color(orange)(f(x))])=color(blue)((color(orange)(ax+1))^2-3)#

#2#. Expand the equation and rewrite in standard form.

#color(blue)(g[color(orange)(f(x))])=color(blue)((color(orange)(ax+1))(color(orange)(ax+1))-3)#

#color(blue)(g[color(orange)(f(x))])=a^2x^2+2ax+1-3#

#color(blue)(g[color(orange)(f(x))])=color(teal)(a^2x^2)+color(purple)(2ax)-2#

#3#. Given that #color(blue)(g[color(orange)(f(x))])=color(teal)(9x^2)+color(purple)(6x)-2#, use the terms in this equation to solve for #a#.

#color(teal)(a^2x^2)=color(teal)(9x^2)color(white)(XXXXXXX)color(purple)(2ax)=color(purple)(6x)#

#a^2=(color(teal)(9x^2))/color(teal)(x^2)color(white)(XXXXXXXx)a=(color(purple)(6x))/(color(purple)(2x))#

#a^2=9color(white)(XXXXXXXXxx)a=3#

#a=+-3#

#4#. However, #color(red)(a!=-3)# because if #color(red)(a=-3)#, then #2ax# should equal #6x#, but it doesn't.

#color(purple)(2ax)=color(purple)(6x)#

#2(color(red)(-3))x=6x#

#-6x!=6xrArr"thus", color(green)(|bar(ul(color(white)(a/a)a=3color(white)(a/a)|)))#

Part B
#1#. Substitute #color(blue)(g(2))# into #color(orange)x# in #color(orange)(f(x))#.

#color(orange)(f[color(blue)(g(2))])=color(orange)(a(color(blue)(x^2-3))+1)#

#2#. Substitute #color(orange)(a=3)# and #color(blue)(x=2)# into the equation.

#color(orange)(f[color(blue)(g(2))])=color(orange)(3(color(blue)((2)^2-3))+1)#

#3#. Solve for #color(orange)(f[color(blue)(g(2))])#.

#color(orange)(f[color(blue)(g(2))])=3(4-3)+1#

#color(orange)(f[color(blue)(g(2))])=3(1)+1#

#color(green)(|bar(ul(color(white)(a/a)f[g(2)]=4color(white)(a/a)|)))#

Part C
#1#. To show that #color(orange)(f[color(blue)(g(x))])=color(blue)(g[color(orange)(f(x))])# has no real solution, set #color(orange)(a(color(blue)(x^2-3))+1)# to equal #color(blue)((color(orange)(ax+1))^2-3)# or #color(brown)(9x^2+6x-2)#. Either way would work, but in this case, we will use #color(blue)((color(orange)(ax+1))^2-3)#.

#color(orange)(a(color(blue)(x^2-3))+1)=color(blue)((color(orange)(ax+1))^2-3)#

#2#. Substitute #a=3#.

#color(orange)(a(color(blue)(x^2-3))+1)=color(blue)((color(orange)(ax+1))^2-3)#

#color(orange)(3(color(blue)(x^2-3))+1)=color(blue)((color(orange)(3x+1))^2-3)#

#3#. Expand the brackets and rewrite the equation in standard form.

#3x^2-9+1=9x^2+6x+1-3#

#3x^2-8=9x^2+6x-2#

#6x^2+6x+6=0#

#4#. Using the quadratic formula, solve for #x#.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(6)+-sqrt((6)^2-4(6)(6)))/(2(6))#

#x=(-6+-sqrt(36-144))/12#

#x=(-6+-sqrt(-108))/12#

#x=(-6+-6sqrt(3)i)/12#

#color(green)(|bar(ul(color(white)(a/a)x=(-1+-sqrt(3)i)/2color(white)(a/a)|)))rArr#since #i=sqrt(-1)#, roots are imaginary (not real)