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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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    The only doubt I had in time of posting is with the understanding of this part"For what values of 'm' is $y=0$,".2011-08-13

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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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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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    with D >= 0, you get$m$>= -5/4. This is the quickest way to the answer; calculus not necessary.2011-08-12
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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.