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If \(n\) is a constant and if there exists a unique value of \(m\) for which the quadratic equation \(x^2+mx+(m+n)=0\) has one real solution, then find \(n\).

 Feb 22, 2019

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 #1
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\(\text{one real solution means the discriminant is 0}\\ \text{for the quadratic }ax^2 + b x + c \text{ the discriminant }\\ D = b^2 - 4ac\\ \text{here we have}\\ D = m^2 - 4(m+n) = 0\\ m^2 -4m - 4n = 0\\ m = 2 \left(1\pm\sqrt{n+1}\right)\)

 

\(\text{we're told that }m \text{ is unique, so }\\ 1+\sqrt{n+1} = 1 - \sqrt{n+1}\\ 2\sqrt{n+1} = 0\\ n=-1\)

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 Feb 23, 2019
 #1
avatar+4449 
+1
Best Answer

\(\text{one real solution means the discriminant is 0}\\ \text{for the quadratic }ax^2 + b x + c \text{ the discriminant }\\ D = b^2 - 4ac\\ \text{here we have}\\ D = m^2 - 4(m+n) = 0\\ m^2 -4m - 4n = 0\\ m = 2 \left(1\pm\sqrt{n+1}\right)\)

 

\(\text{we're told that }m \text{ is unique, so }\\ 1+\sqrt{n+1} = 1 - \sqrt{n+1}\\ 2\sqrt{n+1} = 0\\ n=-1\)

Rom Feb 23, 2019

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