Range of $a$ in $x^2-a=\sqrt{x+a}$ 
The equation $x^2-a=\sqrt{x+a}$ has real or
imaginary roots depending on the values of $a.$
Then range of $a$ for which the equation.
$(a)\;\; $ No real roots
$(b)\;\; $ One real root
$(c)\;\;$ Exactly two real roots
$(d)\;\;$ At least two real roots

what i try
$x^2-a=\sqrt{x+a}\Rightarrow (x^2-a)^2=x+a$
$x^4+a^2-2ax^2=x+a\Rightarrow a^2-(2x^2+1)a+x^4-x=0$
$$a=\frac{(2x^2+1)\pm \sqrt{(2x^2+1)^2-4(x^4-x)}}{2}$$
$$a=\frac{2x^2+1\pm (2x+1)}{2}$$
$$a=x^2-x,a=x^2+x+1$$
$(a)$ For no real roots
$x^2-x-a=0$ and $x^2+x+1-a=0$ has no real roots
So $1+4a>0$ and $\displaystyle 1-4(1-a)>0\Rightarrow a>-\frac{1}{4}.$
How do i solve other parts Help me please

First of all, a quadratic equation has no real roots when its discriminant is negative. So you have $1+4a<0$ and similarly for the other inequality. However, these are not necessarily all the cases. The equation is equivalent to the system
$$\begin{align}
(x^2-a)^2 &= x+a \\
x^2-a &\geq0 \Leftrightarrow x^2\geq a
\end{align}$$
Notice that we don't need $x+a\geq 0$ because that's guaranteed to be non-negative since it's equal to a square. So we're interested in the number of solutions to the system, not just the equation. It might happen that the equation has real roots but they don't satisfy the inequality which would imply the original equation has no real roots. Continuing your work, the equation has roots (by solving the $2$ quadratics)
$$x_{1,2}= \frac{1\pm\sqrt{1+4a}}{2}\\
x_{3,4}=\frac{-1\pm\sqrt{4a-3}}{2}$$
Now you need to plug in each of those in $x^2\geq a$ and solve for $a$. The values for $a$ tell you when the root you plugged in is actually a real root of the original equation. After that, you can determine the answers for (a) to (d).
