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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 $n$.

Let the roots of the quadratic be $r,s.$ Vieta gives $-m=r+s, m+n=rs.$ Thus, $n=rs+r+s \implies n+1=(n+1)(m+1).$ From here, I don't know what to do. I have a feeling that the factoring trick was unnecessary. Any quick (and slick) solutions? Thanks.

Fleccerd
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  • Use the determinant. Twice. – player3236 Sep 03 '20 at 15:59
  • $\left(x+\dfrac m2\right)^2=\dfrac{m^2}{4}-m-n$ has a single solution in $x$ when the RHS is zero. Then $\left(\dfrac m2-1\right)^2=n+1$ has a single solution in $m$ when... –  Dec 21 '21 at 09:58

2 Answers2

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You were on a good path, but apparently overlooked that there is only one real solution for this quadratic equation. So in your calculation, $$ r \ = \ s \ \ \Rightarrow \ \ -m \ = \ 2r \ \ , \ \ m + n \ = \ r^2 \ \ \Rightarrow \ \ m + n \ = \ \left(-\frac{m}{2} \right)^2 \ = \ \frac{m^2}{4} \ \ , $$ which takes you to the same place as finding the discriminant shown in user65203's answer.

We can also get there by considering that a quadratic polynomial with a single zero is a "binomial-square", so $ \ x^2 + bx + c \ = \ \left(x + \frac{b}{2} \right)^2 \ = \ x^2 + bx + \frac{b^2}{4} \ \ . $ For our polynomial then, $ \ b = m \ \Rightarrow \ \frac{m^2}{4} \ = \ m + n \ \ $ (once again). Bringing the terms to one side of the equation produces $ \ m^2 - 4m - 4n \ \ , $ which would be the binomial-square $ \ (m - 2)^2 \ $ if the constant term were $ \ 4 \ \ . $ Hence, $ \ 4n \ = \ -4 \ \ $ (this last bit is just a variant of what user65203 did).

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There is a single real solution when

$$m^2-4(m+n)=0,$$

$$(m-2)^2-4-4n=0$$

and this equation has a single solution in $m$ iff $n=-1$.