1. ## Why are my answers wrong?

1) A quadratic equation in two variables always have two possible solutions in any number system.

My answer: False, it has no solution when a = 0.

2) All equations in two variables, such that both variables have maximum exponent as 2 and minimum as zero represents an equation of a circle.

My answer: False, in equation of a circle, when exponent is 0, (x-h)^0+(y-k)^0=r^2 is not a circle

3) Find the composition of the function as indicated, find domain of composite function and explain your answer(s).

f(x) = ( (4x^2) - 2) / (-4 + 8x)
g(x) = ( (9x^2) -9) / (-12+6x)
Find fog and fof

2. Originally Posted by mwok
1)

3) Find the composition of the function as indicated, find domain of composite function and explain your answer(s).

f(x) = ( (4x^2) - 2) / (-4 + 8x)
g(x) = ( (9x^2) -9) / (-12+6x)
Find fog and fof
$f\left( x \right) = \frac{{4x^2 - 2}}
{{ - 4 + 8x}} = \frac{{2\left( {2x^2 - 1} \right)}}
{{2\left( {4x - 2} \right)}} = \frac{{2x^2 - 1}}
{{4x - 2}} \hfill \\$

$g\left( x \right) = \frac{{9x^2 - 9}}
{{ - 12 + 6x}} = \frac{{3x^2 - 3}}
{{2x - 4}} \hfill \\$

$fog = f\left( {g\left( x \right)} \right) = \frac{{2\left( {\frac{{3x^2 - 3}}
{{2x - 4}}} \right)^2 - 1}}
{{4\left( {\frac{{3x^2 - 3}}
{{2x - 4}}} \right) - 2}} \hfill \\$

$fog = \frac{{2\left( {3x^2 - 3} \right)^2 - \left( {2x - 4} \right)^2 }}
{{4\left( {3x^2 - 3} \right)\left( {2x - 4} \right) - 2\left( {2x - 4} \right)^2 }} \hfill \\$

$= \frac{{2\left( {9x^4 - 18x^2 + 9} \right) - \left( {4x^2 - 16x + 16} \right)}}
{{2\left( {2x - 4} \right)\left[ {2\left( {3x^2 - 3} \right) - \left( {2x - 4} \right)} \right]}} \hfill \\$

$= \frac{{18x^4 - 36x^2 + 18 - 4x^2 + 16x - 16}}
{{2\left( {2x - 4} \right)\left( {6x^2 - 6 - 2x + 4} \right)}} \hfill \\$

$= \frac{{18x^4 - 40x^2 + 16x + 2}}
{{2\left( {2x - 4} \right)\left( {6x^2 - 2x - 2} \right)}} \hfill \\$

$= \frac{{9x^4 - 20x^2 + 8x + 1}}
{{\left( {2x - 4} \right)\left( {6x^2 - 2x - 2} \right)}} \hfill \\$

$= \frac{{9x^4 - 20x^2 + 8x + 1}}
{{4\left( {x - 2} \right)\left( {3x^2 - x - 1} \right)}} \hfill \\$

${\text{For domain, }}x \ne 2,{\text{ and }}3x^2 - x - 1 \ne 0 \hfill \\$

$\Rightarrow x \ne 2,{\text{ and }}x \ne \frac{{1 \pm \sqrt {1^2 - 4\left( 3 \right)\left( { - 1} \right)} }}
{{2\left( 3 \right)}} \hfill \\$

$\Rightarrow x \ne 2,{\text{ and }}x \ne \frac{{1 \pm \sqrt {13} }}
{6} \hfill \\$

${\text{Now, }} \hfill \\$

$fof = f\left( {f\left( x \right)} \right) = \frac{{2\left( {\frac{{2x^2 - 1}}
{{4x - 2}}} \right)^2 - 1}}
{{4\left( {\frac{{2x^2 - 1}}
{{4x - 2}}} \right) - 2}} \hfill \\$

${\text{simplify further}}{\text{.}} \hfill \\$

3. (1) If a = 0, you would no longer have a quadratic which is what the question is asking for.