# Question #0372d

Mar 13, 2016

$a = 3$
$f \left[g \left(2\right)\right] = 4$
imaginary roots$= \frac{- 1 \pm \sqrt{3} i}{2}$

#### Explanation:

Part A
$1$. Given the following, substitute $\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}$ into $\textcolor{b l u e}{x}$ in $\textcolor{b l u e}{g \left(x\right)}$.

$\textcolor{\mathmr{and} a n \ge}{f \left(x\right) = a x + 1}$

$\textcolor{b l u e}{g \left(x\right) = {x}^{2} - 3}$

$\textcolor{b l u e}{g \left[\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}\right]} = \textcolor{b l u e}{{\left(\textcolor{\mathmr{and} a n \ge}{a x + 1}\right)}^{2} - 3}$

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

$\textcolor{b l u e}{g \left[\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}\right]} = \textcolor{b l u e}{\left(\textcolor{\mathmr{and} a n \ge}{a x + 1}\right) \left(\textcolor{\mathmr{and} a n \ge}{a x + 1}\right) - 3}$

$\textcolor{b l u e}{g \left[\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}\right]} = {a}^{2} {x}^{2} + 2 a x + 1 - 3$

$\textcolor{b l u e}{g \left[\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}\right]} = \textcolor{t e a l}{{a}^{2} {x}^{2}} + \textcolor{p u r p \le}{2 a x} - 2$

$3$. Given that $\textcolor{b l u e}{g \left[\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}\right]} = \textcolor{t e a l}{9 {x}^{2}} + \textcolor{p u r p \le}{6 x} - 2$, use the terms in this equation to solve for $a$.

$\textcolor{t e a l}{{a}^{2} {x}^{2}} = \textcolor{t e a l}{9 {x}^{2}} \textcolor{w h i t e}{X X X X X X X} \textcolor{p u r p \le}{2 a x} = \textcolor{p u r p \le}{6 x}$

${a}^{2} = \frac{\textcolor{t e a l}{9 {x}^{2}}}{\textcolor{t e a l}{{x}^{2}}} \textcolor{w h i t e}{X X X X X X X x} a = \frac{\textcolor{p u r p \le}{6 x}}{\textcolor{p u r p \le}{2 x}}$

${a}^{2} = 9 \textcolor{w h i t e}{X X X X X X X X \times} a = 3$

$a = \pm 3$

$4$. However, $\textcolor{red}{a \ne - 3}$ because if $\textcolor{red}{a = - 3}$, then $2 a x$ should equal $6 x$, but it doesn't.

$\textcolor{p u r p \le}{2 a x} = \textcolor{p u r p \le}{6 x}$

$2 \left(\textcolor{red}{- 3}\right) x = 6 x$

$- 6 x \ne 6 x \Rightarrow \text{thus} , \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} a = 3 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Part B
$1$. Substitute $\textcolor{b l u e}{g \left(2\right)}$ into $\textcolor{\mathmr{and} a n \ge}{x}$ in $\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}$.

$\textcolor{\mathmr{and} a n \ge}{f \left[\textcolor{b l u e}{g \left(2\right)}\right]} = \textcolor{\mathmr{and} a n \ge}{a \left(\textcolor{b l u e}{{x}^{2} - 3}\right) + 1}$

$2$. Substitute $\textcolor{\mathmr{and} a n \ge}{a = 3}$ and $\textcolor{b l u e}{x = 2}$ into the equation.

$\textcolor{\mathmr{and} a n \ge}{f \left[\textcolor{b l u e}{g \left(2\right)}\right]} = \textcolor{\mathmr{and} a n \ge}{3 \left(\textcolor{b l u e}{{\left(2\right)}^{2} - 3}\right) + 1}$

$3$. Solve for $\textcolor{\mathmr{and} a n \ge}{f \left[\textcolor{b l u e}{g \left(2\right)}\right]}$.

$\textcolor{\mathmr{and} a n \ge}{f \left[\textcolor{b l u e}{g \left(2\right)}\right]} = 3 \left(4 - 3\right) + 1$

$\textcolor{\mathmr{and} a n \ge}{f \left[\textcolor{b l u e}{g \left(2\right)}\right]} = 3 \left(1\right) + 1$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} f \left[g \left(2\right)\right] = 4 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Part C
$1$. To show that $\textcolor{\mathmr{and} a n \ge}{f \left[\textcolor{b l u e}{g \left(x\right)}\right]} = \textcolor{b l u e}{g \left[\textcolor{\mathmr{and} a n \ge}{f \left(x\right)}\right]}$ has no real solution, set $\textcolor{\mathmr{and} a n \ge}{a \left(\textcolor{b l u e}{{x}^{2} - 3}\right) + 1}$ to equal $\textcolor{b l u e}{{\left(\textcolor{\mathmr{and} a n \ge}{a x + 1}\right)}^{2} - 3}$ or $\textcolor{b r o w n}{9 {x}^{2} + 6 x - 2}$. Either way would work, but in this case, we will use $\textcolor{b l u e}{{\left(\textcolor{\mathmr{and} a n \ge}{a x + 1}\right)}^{2} - 3}$.

$\textcolor{\mathmr{and} a n \ge}{a \left(\textcolor{b l u e}{{x}^{2} - 3}\right) + 1} = \textcolor{b l u e}{{\left(\textcolor{\mathmr{and} a n \ge}{a x + 1}\right)}^{2} - 3}$

$2$. Substitute $a = 3$.

$\textcolor{\mathmr{and} a n \ge}{a \left(\textcolor{b l u e}{{x}^{2} - 3}\right) + 1} = \textcolor{b l u e}{{\left(\textcolor{\mathmr{and} a n \ge}{a x + 1}\right)}^{2} - 3}$

$\textcolor{\mathmr{and} a n \ge}{3 \left(\textcolor{b l u e}{{x}^{2} - 3}\right) + 1} = \textcolor{b l u e}{{\left(\textcolor{\mathmr{and} a n \ge}{3 x + 1}\right)}^{2} - 3}$

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

$3 {x}^{2} - 9 + 1 = 9 {x}^{2} + 6 x + 1 - 3$

$3 {x}^{2} - 8 = 9 {x}^{2} + 6 x - 2$

$6 {x}^{2} + 6 x + 6 = 0$

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(6\right) \pm \sqrt{{\left(6\right)}^{2} - 4 \left(6\right) \left(6\right)}}{2 \left(6\right)}$

$x = \frac{- 6 \pm \sqrt{36 - 144}}{12}$

$x = \frac{- 6 \pm \sqrt{- 108}}{12}$

$x = \frac{- 6 \pm 6 \sqrt{3} i}{12}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{- 1 \pm \sqrt{3} i}{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \Rightarrow$since $i = \sqrt{- 1}$, roots are imaginary (not real)