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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.

 Jan 6, 2019

Best Answer 

 #2
avatar+701 
+1

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

 #1
avatar+4467 
+3

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

 

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

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 Jan 6, 2019
edited by Rom  Jan 6, 2019
 #3
avatar+701 
+4

I am pretty sure you factored \(c^2 - 2c + 1 = 0\) incorrectly. 

 #2
avatar+701 
+1
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

 #4
avatar+98184 
+2

Good job, PM...!!!!

 

cool cool cool

CPhill  Jan 6, 2019

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