Question

If there are two real solutions to the equation `x^2+4x+c=0`, what is a possible value of `c`?

Answer 1

`x=+3" " ;" " x=-7`

Explanation:

Bear in mind that `c` is the y-intercept.

First tray failed! I thought 2+2=4 so try `(x+2)(x+2)` the x-axis was tangential to the minimum (1 value). The question definitely states 2 solutions.

The next move: Difference of 4; 2 did not work so if I made one of them 3 the other one to give a difference of 4 is 7. So we need -3 and +7. That worked

`(x-3)(x+7)" " =" " x^2+7x-3x-21 " "`

`= x^2+4x-21`

So at least one of the solution sets is `x=+3 ; x=-7`