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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

#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
+5088
+1
$$\text{one real solution means the discriminant is 0}\\ \text{for the quadratic }ax^2 + b x + c \text{ the discriminant }\\ D = b^2 - 4ac\\ \text{here we have}\\ D = m^2 - 4(m+n) = 0\\ m^2 -4m - 4n = 0\\ m = 2 \left(1\pm\sqrt{n+1}\right)$$
$$\text{we're told that }m \text{ is unique, so }\\ 1+\sqrt{n+1} = 1 - \sqrt{n+1}\\ 2\sqrt{n+1} = 0\\ n=-1$$