# How do you find the c that makes the trinomial x^2+9x+c a perfect square?

Apr 30, 2017

$c = \frac{81}{4}$

#### Explanation:

Note that:

${\left(x + \frac{b}{2}\right)}^{2} = {x}^{2} + b x + {b}^{2} / 4$

So if $b = 9$ then:

$c = {b}^{2} / 4 = {\textcolor{b l u e}{9}}^{2} / 4 = \frac{81}{4}$

$\textcolor{w h i t e}{}$
Bonus

More generally, we find:

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

which if $a = 1$ simplifies to:

${x}^{2} + b x + c = {\left(x + \frac{b}{2}\right)}^{2} + \left(c - {b}^{2} / 4\right)$

as in our example.

Apr 30, 2017

The pattern is:

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

Match the terms of the given equation with the right side of the pattern.

#### Explanation:

Match the terms of the given equation, ${x}^{2} + 9 x + c$ with the right side of the pattern.

Matching the first terms:

${x}^{2} = {x}^{2}$ does not help us.

Matching the second terms gives us the following equation:

$2 a x = 9 x$

We can use the above equation to solve for the value of "a" by dividing both sides of the equation by 2x:

$a = \frac{9}{2}$

Matching the constant terms, gives us the following equation:

${a}^{2} = c$

Substitute $\frac{9}{2}$ for "a":

${\left(\frac{9}{2}\right)}^{2} = c$

$c = \frac{81}{4}$