# How do you find the value of "a" such that t^2 + 8t +a is a perfect square?

Apr 22, 2015

You can use the algebraic identity

${\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2}$

If you compared these two expressions, you'll notice that

${t}^{2} + {\underbrace{8 t}}_{\textcolor{b l u e}{2 x y}} + {\overbrace{a}}^{\textcolor{b l u e}{{y}^{2}}}$

Since, in your case, $x$ will be equal to $t$, you'll get

$\stackrel{4}{\cancel{8}} \cdot \cancel{t} = \cancel{2} \cdot \cancel{t} \cdot y \implies y = 4$

and

$a = {y}^{2} \implies \textcolor{g r e e n}{a} = {4}^{2} = \textcolor{g r e e n}{16}$

${t}^{2} + 8 t + 16 = {t}^{2} + 2 \cdot t \cdot 4 + {4}^{2} = {\left(t + 4\right)}^{2}$

Apr 22, 2015

We can apply the identity color(blue)((a+b)^2 = a^2 + 2ab + b^2

The middle term $2 a b$ is 2 * First Term(a) * Second Term(b)

So we can write the middle term of '${t}^{2} + \textcolor{red}{8 t} + a$' as '$\textcolor{red}{2 \cdot t \cdot 4}$'

It tells us that the first term is 't' and the second one is '4'

${t}^{2} + 8 t + {4}^{2}$ would make it a Perfect Square ${\left(t + 4\right)}^{2}$

Hence color(green)( a = 4^2 = 16