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# help

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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

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

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

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Jan 6, 2019
edited by Rom  Jan 6, 2019
#3
+701
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I am pretty sure you factored $$c^2 - 2c + 1 = 0$$ incorrectly.

#2
+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

#4
+98184
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Good job, PM...!!!!

CPhill  Jan 6, 2019