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 $c$.

 Jun 3, 2018

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,



 Jun 3, 2018

10 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.