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.

grif389 Jan 6, 2019

#2**+5 **

If the polynomial has 1 real root, then the discriminant mut equal 0. We can rewrite the polynomial as \(x^2 + cx + x + c\).

The discriminant is always \(b^2-4ac\), and setting a = 1 and b = c+1, we have \((c+1)^2 - 4c = 0\). Simplifying, we have \(c^2 + 2c +1 - 4c = 0 \Rightarrow c^2 - 2c +1 = 0\). Factoring, we have \((c-1)^2 = 0\). Taking the square root of both sides, we have \(c -1 = 0 \Rightarrow \boxed{c = 1}\).

There is only one solution since there is only once real root. The product of all possible values of c is 1.

Hope this helps,

- PM

PartialMathematician Jan 6, 2019

#1**+2 **

\(\text{1 real root indicates that the Discriminate }D = b^2 - 4ac \text{ is equal to }0\\ D=(c+1)^2 - 4(1)(c) = \\ c^2 + 2c + 1 - 4c =\\ c^2 - 2c+1 = 0\)

\((c-1)^2 = 0\\ c=1\)

.Rom Jan 6, 2019

#3

#2**+5 **

Best Answer

If the polynomial has 1 real root, then the discriminant mut equal 0. We can rewrite the polynomial as \(x^2 + cx + x + c\).

The discriminant is always \(b^2-4ac\), and setting a = 1 and b = c+1, we have \((c+1)^2 - 4c = 0\). Simplifying, we have \(c^2 + 2c +1 - 4c = 0 \Rightarrow c^2 - 2c +1 = 0\). Factoring, we have \((c-1)^2 = 0\). Taking the square root of both sides, we have \(c -1 = 0 \Rightarrow \boxed{c = 1}\).

There is only one solution since there is only once real root. The product of all possible values of c is 1.

Hope this helps,

- PM

PartialMathematician Jan 6, 2019