# How do you find the missing number or term to make a perfect square trinomial. x^2+1/3x+___ ?

Jun 13, 2015

$\frac{1}{36}$

#### Explanation:

you're looking for a square trinomial, which is of the form

${x}^{2} + 2 \alpha x + {\alpha}^{2}$

Let's find ${\alpha}^{2}$

${x}^{2} + \frac{1}{3} x + {\alpha}^{2} = {x}^{2} + 2 \alpha x + {\alpha}^{2}$

Which means $\frac{1}{3} = 2 \alpha \implies \alpha = \frac{1}{6}$

In fact, ${\left(x + \frac{1}{6}\right)}^{2} = {x}^{2} + 2 \frac{1}{6} x + \frac{1}{36} = {x}^{2} + \frac{1}{3} x + \frac{1}{36}$

So ${\alpha}^{2} = \frac{1}{36}$

Jun 13, 2015

Perfect square trinomials are of the form

${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$.

In this case $a = x$, so $2 b = \frac{1}{3}$, so $b = \frac{1}{6}$ and ${b}^{2} = \frac{1}{36}$

#### Explanation:

Compare:

${x}^{2} + \frac{1}{3} x +$_

with

${a}^{2} + 2 a b + {b}^{2}$

We can obviously put $a = x$, giving

${x}^{2} + 2 b x + {b}^{2}$

Comparing the coefficient of $x$, we have

$2 b = \frac{1}{3}$

Divide both sides by $2$ to get:

$b = \frac{1}{6}$

So ${b}^{2} = {\left(\frac{1}{6}\right)}^{2} = \frac{1}{36}$

which is the missing term.