Question

If $g\left(x\right)=3x+\sqrt{x}$, find $g(d^2+6d+9)$.

Answer 1

$3d^2+19d+30$ or $3d^2+17d+24$:
$g(d^2+6d+9)=3(d^2+6d+9)+\sqrt{d^2+6d+9}$
$=3d^2+18d+27+\sqrt{{(d+3)}^2}$
$=3d^2+18d+27+d+3$ OR $3d^2+18d+27-(d+3)$
$=3d^2+19d+30$ OR $3d^2+17d+24$
(if $d+3>0$) (if $d+3<0$)