Question

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

Answer 1

`c=49/4`

Explanation:

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

Answer 2

`c=49/4=12 1/4`

Explanation:

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

In `x^2-7x+c`, determinant `(-7)^2-4xxc=0` means

`49-4c=0`

i.e. `4c=49`

or `c=49/4=12 1/4`,

and then `x^2-7x+49/4=(x-7/2)^2`