Question

Find all positive integer values of $c$ such that the equation $x^2-7x+c=0$ only has roots that are real and rational. Express them in decreasing order, separated by commas.

Answer 1

12, 10, 6 (descending order)

Explanation:

`ax^2+bx+c=0` has real and rational roots if
`b^2-4ac` is a nonnegative perfect square.
`b^2-4ac` is called the discriminant.

Let's calculate the discriminant of
`x^2-7x+c=0`.

`b^2-4ac`

`(-7)^2-4(1)(c)`

`49-4c`

We want
`49-4c` to be be positive or a zero perfect square.

So let's first solve the inequality:

`49-4c \ge 0`

Add
`4c` on both sides:

`49 \ge 4c`

`4c \le 49`

Divide both sides be 4:

`c \le (49)/(4)`

`c \le 12.25`.

We know
`c` has to be a positive integer.

So we only need to evaluate
`49-4c` for 12 numbers.

`c=1` returns
`49-4(1)=49-4=45` which is not a perfect square.

`c=2` returns
`49-4(2)=49-8=41` which is not a perfect square.

`c=3` returns
`49-4(3)=49-12=37` which is not a perfect square.

`c=4` returns
`49-4(4)=49-16=33` which is not a perfect square.

`c=5` returns
`49-4(5)=49-20=29` which is not a perfect square.

`c=6` returns
`49-4(6)=49-24=25` which is a perfect square.

`c=7` returns
`49-4(7)=49-28=21` which is not a perfect square.

`c=8` returns
`49-4(8)=49-32=17` which is not a perfect square.

`c=9` returns
`49-4(9)=49-36=13` which is not a perfect square.

`c=10` returns
`49-4(10)=49-40=9` which is a perfect square.

`c=11` returns
`49-4(11)=49-44=5` which is not a perfect square.

`c=12` returns
`49-4(12)=49-48=1` which is a perfect square.