Question

Question #0372d

Answer 1

`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)