# How do you find the c that makes the trinomial x^2-7x+c a perfect square?

Aug 19, 2017

$c = \frac{49}{4}$

#### Explanation:

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$
In the above see the relation between middle term and the last term: What do you need to do to $- 2 a b$ to become ${b}^{2}$?
divide by $2 a$ then square the results:
$\frac{- 2 a b}{2 a} = - b$ => square: ${b}^{2}$
Now in this case:
${x}^{2} - 7 x + c$
$c = {\left(\frac{7}{2}\right)}^{2} = \frac{49}{4}$

Aug 19, 2017

$c = \frac{49}{4} = 12 \frac{1}{4}$

#### Explanation:

If a trinomial of the form ${x}^{2} + b x + c$ is a perfect square, then its determinant ${b}^{2} - 4 c$ must be equal to zero.

In ${x}^{2} - 7 x + c$, determinant ${\left(- 7\right)}^{2} - 4 \times c = 0$ means

$49 - 4 c = 0$

i.e. $4 c = 49$

or $c = \frac{49}{4} = 12 \frac{1}{4}$,

and then ${x}^{2} - 7 x + \frac{49}{4} = {\left(x - \frac{7}{2}\right)}^{2}$