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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 $c$.

Lightning  Jun 3, 2018
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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

GYanggg  Jun 3, 2018

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