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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 $n$.

 Aug 3, 2020
 #1
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Since this is a monic quadratic, we can use vieta's formula : https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas.

So, we see that mn=n+m, and -m-n=m. Now, here is where is get's a bit tricky. When we add m to both sides of the second equation, we see that -n=2m. Divide by -1, we see that n=-2m.

 

Here is what I want you do to: plug -2m for n in the first equation, then solve for m. 

 

Then, put that into n=-2m, and solve for n.

 

Good luck!

 

:)

 Aug 3, 2020

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