If the equation $|x^2+bx+c|=k$ has four real roots then which of the following option is correct :
(a) $b^2-4c >0$ and $0<k<\frac{4c-b^2}{4}$
(b) $b^2-4c <0$ and $0<k<\frac{4c-b^2}{4}$
(c) $b^2-4c >0$ and $0<k>\frac{4c-b^2}{4}$
Please suggest how to proceed in this problem , I am getting no clue on this.. please help thanks.
Best Answer
Let's narrow the three possibilities to one, and confirm whether it actually works.
First, note that there are three possibilities for a graph of $y=x^2+bx+c$: It passes above the the $x$-axis (and has no real roots), it just 'kisses' the $x$-axis (and has one real root), or it crosses the axis twice (two real roots). In the first two cases, the quadratic is always positive and so the absolute value has no effect. But, in the third case, the portion of the parabola below the $y$-axis is reflected into the $x>0$ half plane.
Suppose we now cross these graphs with a horizontal line $y=k$. If we have a parabola, then this line can cross the parabola at most twice; since we want four, we must have the two real roots of $x^2+bx+c=0$. If we recall the quadratic formula, this requires that $b^2-4c>0$ so that the plus/minus gives two real solutions. That eliminates option (b).
If we do have $b^2-4c>0$, then we can choose a range of $k$ such that it crosses the reflected parabola twice. By visual inspection of the reflected parabola, it should be of the form $0<k<k_{max}$. Only option (a) is of this form, with $k_{max}=\frac{4c-b^2}{4}$. To check this, note that the most negative value that $x^2+bx+c=(x+\frac{b}{2})^2+\frac{4c-b^2}{4}$ can express is $\frac{4c-b^2}{4}$ (which is negative since we require $b^2-4c>0$.) Upon taking the absolute value, this means that the reflected 'hump' of the parabola has a height of $\frac{4c-b^2}{4}$ which is indeed $k_{max}$ as in answer (a).