+1245 If the polynomial $x^2+bx+c$ has exactly one real root and $b=c+1$, find the value of the product of all possible values of $c$.
+983 Hey lightning!
If a quadratic equation has exxactly one real root, the discriminant is equal to zero.
In the quadratic equation: \(ax^2+bx+c\),
The discriminant is: \(b^2-4ac\).
In this specific problem, the discriminant is \(b^2-4c\), since \(a=1\).
From the information given, we can set up the systems:
\(b^2-4c=0,\\ b=c+1.\)
Solving the systems, we can substitute:
\((c+1)^2-4c=0\\ c=1\)
I hope this helped,
Gavin
