Mathematics - Quadratic Equation Question with Solution | TestHub

Let $t=x^2+3x$. Completing the square, $t=(x+3/2{)}^2-9/4$. Since $(x+3/2{)}^2\ge 0$, we have $t\ge -9/4$. The given expression becomes $g(t)=(t+2)(t+4)+5=t^2+6t+8+5=t^2+6t+13$. This is a quadratic in $t$ with its vertex at $t=-6/(2\cdot 1)=-3$. Since the parabola opens upwards and its vertex is at $t=-3$, which is less than the allowed domain $t\ge -9/4$, the function $g(t)$ is increasing for $t\ge -9/4$. Therefore, the minimum value of $g(t)$ occurs at $t=-9/4$. Substituting $t=-9/4$ into $g(t)$: $g(-9/4)=(-9/4{)}^2+6(-9/4)+13=81/16-54/4+13=81/16-216/16+208/16=(81-216+208)/16=73/16$.