# limits of $x$ for $( x-a)(x-b)>c$

After solving a quadratic inequality, I've ended up getting the solution in the format-

$(x-a)(x-b)>c$,

where $a,b,c$ are real constants, then how can I decide on the limits of $x$. will the limits depend upon $a,b$ alone?

Not sure if I understand the question correctly, the general approach of solving quadratic inequality is to rearrange the inequality to the form of $(x-a)(x-b)>0$ or $(x-a)(x-b)<0$. The solution of the former one is $x>\max(a,b)$ $\cup$ $x<\min(a,b)$. The solution of the later one is $a<x<b$ or $b<x<a$ depends on which one is larger.
If the quadratic polynomial cannot be factorized(in $R[x]$) into two linear polynomial than the solution of the inequality is either $R$ or $\phi$.
• The solution to the former also has the branch $x<\min(a,b)$. Feb 7 '15 at 10:05
the global minimum of $f(x) = (x-a)(x-b)$ is $m = f(\frac{a+b}{2}) = -\frac{(a-b)^2}{4}$ now you have to consider two cases:
(a) if $c < m,$ then $(x-a)(x-b) > c$ is true for all $x.$
(b) case $c \ge m.$ let the roots of $f(x) = c$ be $x= x_1, x_2$ and $x_1 < x_2.$ now $(x-a)(x-b) > c$ is true for all $x< x_1$ or $x > x_2.$