Answer:
[tex]n^2 + \frac{2n}{3} + \frac{1}{9} = \frac{1}{9}[/tex] -- Perfect square trinomial
[tex](n + \frac{1}{3})^2 = \frac{1}{9}[/tex] ---Binomial squared
Step-by-step explanation:
Given
[tex]n^2 + \frac{2n}{3}[/tex]
Solving (a): Perfect square trinomial
We have:
[tex]n^2 + \frac{2n}{3}[/tex]
Express as an equation
[tex]n^2 + \frac{2n}{3} =[/tex]
Start----------------------------
Take coefficient of n i.e. (2/3)
Half it: i.e. (1/3)
Square it: (1/63^2
Add to both sides of the equation
---------------------------End
So, we have:
[tex]n^2 + \frac{2n}{3} + (\frac{1}{3})^2 = (\frac{1}{3})^2[/tex]
Remove brackets
[tex]n^2 + \frac{2n}{3} + \frac{1}{9} = \frac{1}{9}[/tex]
Solving (b): Binomial Squared
[tex]n^2 + \frac{2n}{3} + \frac{1}{9} = \frac{1}{9}[/tex]
Expand
[tex]n^2 + \frac{n}{3}+ \frac{n}{3} + \frac{1}{9} = \frac{1}{9}[/tex]
Factorize:
[tex]n(n + \frac{1}{3})+ \frac{1}{3}(n + \frac{1}{3}) = \frac{1}{9}[/tex]
Factor out n + 1/3
[tex](n + \frac{1}{3})(n + \frac{1}{3}) = \frac{1}{9}[/tex]
Express as square
[tex](n + \frac{1}{3})^2 = \frac{1}{9}[/tex]
