# When does this quadratic equation have a real solution?

While solving some quantitative problem,I got stuck in this one:

For what values of $m$ is $y = 0$, if $y=x^2+(2m+ 1)x+m^2-1$? ($x$ is a real number).

Could anybody help me to understand this question?

EDIT [Srivatsan]: Based on what has been agreed upon as the intended question, I have improved the title.

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Imagine how the main page would look if everyone called their question something like "Explain this question"... – joriki Aug 12 '11 at 11:40
@joriki:Fixed. $\quad$ – Quixotic Aug 12 '11 at 14:13
Thanks. An intermediate level of detail like "Solutions of a quadratic equation" might be preferable, though, or perhaps, if you don't find that specific enough (or didn't know when you asked that this was a quadratic equation), a more succinct version like "Solutions of $x^2+(2m+1)x+m^2-1=0$". I suggest you look at the main page, compare how this title looks there in comparison to those around it, and, again, imagine how the main page would look if they all looked like this one. – joriki Aug 12 '11 at 15:15
The only doubt I had in time of posting is with the understanding of this part"For what values of 'm' is $y=0$,". – Quixotic Aug 13 '11 at 6:31

I’m assuming that $x$ is given, and you want to find $m$ so that $x^2 + (2m+1)x + m^2 - 1 = 0$. To do this, just treat the equation as a quadratic in $m$, with $x$ as a constant, and rewrite it as $m^2 + 2xm + (x^2 +x - 1) = 0$. The quadratic formula now gives \begin{align*}m &= \frac12\left(-2x\pm\sqrt{4x^2 - 4(x^2+x-1)} \right)\\ &= -x\pm\sqrt{1-x} \end{align*}. Alternatively, it’s easy to complete the square: $m^2 + 2xm + (x^2 +x - 1) = (m+x)^2 + x - 1$, which gets you to $m = -x \pm \sqrt{1-x}$ even faster.

Added: It occurs to me that what’s wanted might be the range of values of $m$ for which there is a real $x$ making $y=0$. Let $f(x) = -x + \sqrt{1-x}$ and $g(x) = -x - \sqrt{1-x}$. Clearly these are defined only for $x\le 1$. Now $f'(x) = -1 - \frac{1}{2\sqrt{1-x}}$ , so $f'(x)<-1$ for all $x<1$; this implies that $f(x)$ attains its minimum at $x=1$, where $f(1) = -1$, and that $\lim\limits_{x\to -\infty}f(x) = \infty$. Thus, the set of possible values of $m$ includes at least $[1,\infty)$.

$g'(x) = -1 + \frac{1}{2\sqrt{1-x}} = 0$ when $\sqrt{1-x} = \frac12$, i.e., when $x = \frac34$, and it’s easy to check that $g(x)$ has a minimum of $-\frac54$ at this point. Thus, the set of possible values of $m$ is actually $\left[-\frac54,\infty\right)$.

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+1,Beautiful explanation,thanks :-) – Quixotic Aug 12 '11 at 10:28

Look at the discriminant which should be $0$. So you have

\begin{align*} 0 &= (2m+1)^{2} - 4 \cdot (m^{2}-1) \\ &= 4m^{2} + 4m +1 - 4m^{2} +4 \\ &= 4m+5 \end{align*}

which gives $m= -\frac{5}{4}$

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Hm,I did think about this,but what exactly the question meant here "For what values of 'm' is $y=0$"?Do you mean that for any real $x$ and $m= -\frac{5}{4} \Rightarrow y = 0$? – Quixotic Aug 12 '11 at 8:48
This just finds the value of $m$ for which the equation has a double root. When $m=-\frac54$ you have $y=\left(x-\frac34\right)^2$, so $y=0$ when and only when $x=\frac34$. – Brian M. Scott Aug 12 '11 at 9:27
The discriminant has to be $\geq 0$. – Eelvex Aug 12 '11 at 9:27
with D >= 0, you get m >= -5/4. This is the quickest way to the answer; calculus not necessary. – David Lewis Aug 12 '11 at 11:45

The parametrized quadratic $\rm\: x^2 + B(m)\ x + C(m)\:$ has real roots precisely where its discriminant $\rm\: D = B^2 - 4\:C \ge 0\:.\:$ This will be a single $\rm\:m\:$ interval when the discriminant is linear in $\rm\:m\:.\:$ For example, for $\rm\: B = 2\:a\ m + b,\ \ C = a^2\:m^2+c\:,\:$ we find $\rm\:D = 4\:a\:b\ m + b^2 -4\:c\:$ so it has real roots iff $\rm\:m \ge (4\:c-b^2)/(4\:a\:b)\:.\:$ Your special case is $\rm\:a = b = 1,\ c = -1\:$ so $\rm\:m \ge -5/4\:.$ Note that I have ignored the "degenerate" cases $\rm\:a=0\:$ or $\rm\:b = 0\:$ (which do not apply in your case).

For some related questions see here and here.

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