Question

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

Answer 1

You can use the algebraic identity

`(x + y)^(2) = x^2 + 2xy + y^2`

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

`t^2 + underbrace(8t)_(color(blue)(2xy)) + overbrace(a)^(color(blue)(y^2))`

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

`stackrel(4)(cancel(8)) * cancel(t) = cancel(2) * cancel(t) * y => y = 4`

and

`a = y^2 => color(green)(a) = 4^2 = color(green)(16)`

Your expression will become

`t^2 + 8t + 16 = t^2 + 2 * t * 4 + 4^2 = (t + 4)^2`

Answer 2

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

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

So we can write the middle term of '`t^2 + color(red)(8t) +a`' as '`color(red)(2*t*4)`'

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

`t^2 + 8t + 4^2` would make it a Perfect Square `(t+4)^2`

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