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

 Aug 17, 2019
 #1
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Help me please I need help quick...

 Aug 17, 2019
 #2
avatar+1342 
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\(If\ n\ is\ a \ constant\ and\ if\ there\ exists\ a\ unique\ value\ of\ m\ for \\ which\ the\ quadratic\ equation\ x^2 + mx + (m+n) = 0\ has\ one \\ real\ solution,\ then\ find\ n. \)

 

Sorry The not being able to see the whole problem at once bothered me.

 Aug 17, 2019
 #3
avatar+8519 
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\({If \ n\ is\ a\ constant\ and\ if\ there\ exists \\ a\ unique\ value\ of\ m\ for\ which\ the\ quadratic\ equation\\ x^2 + mx + (m+n) = 0\ has\ one\ real\ solution,\ then\ find\ n. } \)

 

\( \color{BrickRed}x^2 + mx + (m+n) = 0\\ x=-\frac{m}{2}\pm\sqrt{(\frac{m}{2})^2-m-n}\ |(p-q-formula)|\)

\((\frac{m}{2})^2-m-n\ge 0\\ n \le\frac{m^2}{4}-m\\ \)

 

\(n\in\mathbb{R}\ |\ n\le\frac{m^2}{4}-m\)

laugh  !

 Aug 17, 2019
edited by asinus  Aug 17, 2019

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