Question

Find for what value of $a$, equation $x^2+2x-(a^3-3a-3)=0$ has real roots.

Answer 1

$x^2+2x-(a^3-3a-3)=0$

For the equation to have real roots $△\ge 0$

$b^2-4ac\ge 0$

$4+4(a^3-3a-3)\ge 0$

$1+a^3-3a-3\ge 0$

$a^3-3a-2\ge 0$

For $a=-1△=0$

For other two roots,

(Image)

$a^2-a-2=0$

$a^2-2a+a-2=0$

$a(a-2)+1(a-2)=0$

$a=-1,2$

$\therefore$Values of a are $-1,2$