Question

plz help thanks so much if you help

Answer 1

Answer:

$n^2 + \frac{2n}{3} + \frac{1}{9} = \frac{1}{9}$ -- Perfect square trinomial

$(n + \frac{1}{3})^2 = \frac{1}{9}$ ---Binomial squared

Step-by-step explanation:

Given

$n^2 + \frac{2n}{3}$

Solving (a): Perfect square trinomial

We have:

$n^2 + \frac{2n}{3}$

Express as an equation

$n^2 + \frac{2n}{3} =$

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:

$n^2 + \frac{2n}{3} + (\frac{1}{3})^2 = (\frac{1}{3})^2$

Remove brackets

$n^2 + \frac{2n}{3} + \frac{1}{9} = \frac{1}{9}$

Solving (b): Binomial Squared

$n^2 + \frac{2n}{3} + \frac{1}{9} = \frac{1}{9}$

Expand

$n^2 + \frac{n}{3}+ \frac{n}{3} + \frac{1}{9} = \frac{1}{9}$

Factorize:

$n(n + \frac{1}{3})+ \frac{1}{3}(n + \frac{1}{3}) = \frac{1}{9}$

Factor out n + 1/3

$(n + \frac{1}{3})(n + \frac{1}{3}) = \frac{1}{9}$

Express as square

$(n + \frac{1}{3})^2 = \frac{1}{9}$