Answer
Since the length must be a positive number, the length is $3$
Work Step by Step
Let's consider a trinomial in this form: $~x^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$
Then the next step is to rewrite the trinomial as follows:
$~x^2+bx+c = (x+r)~(x+s)$
We can rewrite the given equation:
$a^2 +(a+1)^2 = 25$
$a^2 +(a^2+2a+1) = 25$
$2a^2 +2a+1 = 25$
$2a^2 +2a-24=0$
$(2)~(a^2 +a-12)=0$
To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = 1~$ and $~r\times s = -12$. We can see that $(4)+(-3) = 1~$ and $(4)\times (-3) = -12$
We can solve the equation as follows:
$a^2 +(a+1)^2 = 25$
$a^2 +(a^2+2a+1) = 25$
$2a^2 +2a+1 = 25$
$2a^2 +2a-24=0$
$(2)~(a^2 +a-12)=0$
$(2)~(a+4)~(a-3)=0$
$a+4=0~~$or $~~a-3=0$
$a = -4~~$ or $~~a=3$
Since the length must be a positive number, the length is $3$
