Question

Solve the equation x |x + 1| + a = 0 for every real number a.

Answer 1

We consider two cases :
(a) x < -1, (b) x $\ge$-1
In case (a), the equation reduces to the form
x(- x - 1) + a = 0
or x${}^2$ + 1 - a = 0 ............(1)
We are interested in finding real roots of (1) which satisfy the condition (a).
Now the condition for (a) to have real roots is
$1+4a\ge 0$ or $a\ge -\frac{1}{4}$
[Note that for real roots, the discriminant of (1) must be non - negative ].
The roots of (1) are
$x_1=\frac{-1+\sqrt{1+4a}}{2}$,$x_1=\frac{-1-\sqrt{1+4a}}{2}$
Out of these roots, we have to take those which satisfy the condition x < -1.
To do this, we have solve the inequalities
$x_1=\frac{-1+\sqrt{1+4a}}{2}<-1$
and $x_1=\frac{-1-\sqrt{1+4a}}{2}<-1$
The first inequality reduces to $1+\sqrt{1+4a}<0$,
Which does not hold for values of $a\ge -\frac{1}{4}$
The second inequality reduces to $\sqrt{1+4a}>1$ and is valid for a > 0 as can be easily seen.
Hence, for a > 0, the original equation has one real root at $x=\frac{-1-\sqrt{1+4a}}{2}$
that satisfies the condition x < -1 and for $a\le 0$, it has no such root.
In case (b), the given equation takes the form
$x^2$ + x + 1 = 0
Discussing as in case (a), we will see that the given equation has two real roots for $0\le a\le \frac{1}{4}$.
$\frac{-1+\sqrt{1-4a}}{2}$ and $\frac{-1-\sqrt{1-4a}}{2}$
For a < 0, the equation has the root at
$x=\frac{-1+\sqrt{1-4a}}{2}$
For $a>\frac{1}{4}$, the equations has no roots in the domain x > -1.
Combining the cases (a) and (b) , we get the final answer .
For a < 0, $x=\frac{-1+\sqrt{1-4a}}{2}$
For $0\le a\le \frac{1}{4},x=\frac{-1-\sqrt{1-4a}}{2},\frac{-1+\sqrt{1-4a}}{2}$
and for $a>\frac{1}{4}$, the equation has no root