pistolpete wrote:Problem #1
x^2 - 2x - 15=0
If you're not sure, check! . . . . .Factor the quadratic
:. . . . .
Then solve the factors:. . . . .. . . . .
Your quadratic does indeed have the required zeroes, so it must be correct!
pistolpete wrote:Problem #2
x^2 - 5 1/2x - 3=0
It would almost certainly be better to use an improper fraction, rather than a mixed number, for the coefficient of the linear term. So I would suggest changing the "five and a half" to "eleven halves".. . . . .
To check, use the Quadratic Formula
, or else multiply through to clear the denominator and then factor. I prefer the latter:. . . . .. . . . .. . . . .. . . . .
Completing the solution, you should get that x = 6 or x = -1/2, as required, so your quadratic is a valid solution.