x² + 9x + 1
let's start by grouping the ones with the same variable
(x² + 9x) + 1
(x² + 9x + [?]²) + 1
so, we seem to be missing a value there, to get a perfect square trinomial, let's recall a perfect square trinomial has a middle term that is 2 * "other two values", namely
[tex]\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2[/tex]
so, the middle term on this group will be 9x.
we know 2*x*[?] = 9x, so then
[tex]\bf 2\cdot x\cdot \boxed{?}=9x\implies \boxed{?}=\cfrac{9x}{2x}\implies \boxed{?}=\cfrac{9}{2}[/tex]
so that's our mystery felllow.
now, let's bear in mind that we'll be borrowing from our very good friend Mr Zero, 0, so if we add (9/2)², we also have to subtract (9/2)².
[tex]\bf \left[x^2+9x+\left( \cfrac{9}{2} \right)^2 - \left( \cfrac{9}{2} \right)^2 \right]+1\implies \left[x^2+9x+\left( \cfrac{9}{2} \right)^2 \right]+1- \left( \cfrac{9}{2} \right)^2 \\\\\\ \left( x+\cfrac{9}{2} \right)^2+1-\cfrac{81}{4}\implies \left( x+\cfrac{9}{2} \right)^2-\cfrac{77}{4}[/tex]
