Find the value of the product of roots of this quadratic equation It is given...

Step 1
Without applying the factorization of a "difference of two squares" or Viete's relations, we can still use the information stated in the problem. If we call the two roots of the quadratic equation r and -r , then we have
${\displaystyle r^2+(p+3)\times r-p^2=0}$
and
${\displaystyle {[-r]}^2+(p+3)\times [-r]-p^2}$
${\displaystyle =r^2-(p+3)\times r-p^2=0}$
This means that
${\displaystyle r^2=-(p+3)\times r+p^2}$
${\displaystyle =(p+3)\times r+p^2\Rightarrow 2\times (p+3)\times r=0}$
So either ${\displaystyle r=0}$ or ${\displaystyle p=-3}$
But if ${\displaystyle r=0=-r}$, then
${\displaystyle 0^2+(p+3)\times 0-p^2=0}$ would reqiure ${\displaystyle p=0}$,
which would then make the quadratic equation
${\displaystyle x^3+3\times x=0}$
But that polynomial factors as
${\displaystyle x\times (x+3)=0}$,
so we couldn't have both roots equal to zero.
Instead, it must be that ${\displaystyle p=-3}$, making the equation
${\displaystyle x^2+0\times x-{(-3)}^2=x^2-9=0}$,
for which the roots are given by
${\displaystyle r^2=9\Rightarrow r=+3,-3;}$
the product of the roots is thus -9
Step 2
Another way to arrive at this conclusion is that
${\displaystyle y=x^2+(p+3)\times x-p^2}$
is the equation of an "upward-opening" parabola, for which we want the x-intercepts to be ${\displaystyle x=-r}$ and ${\displaystyle x=r}$. Its axis of symmetry is located midway between these intercepts, so we have ${\displaystyle h=0}$ in the "vertex form" of the parabola's equation,
${\displaystyle y={(x-0)}^2-p^2}$
(The vertex is definitely "below" the x-axis at ${\displaystyle (0,-p^2)}$, so we know these x-intercepts exist.) The equation of the parabola is therefore ${\displaystyle y=x^2-p^2}$, making ${\displaystyle p+3=0}$ and the rest of the argument above follows.

Step 1
We know that this equation has at most two roots in the set of reals.
Let's denote them with a and -a.
Then your equation is
${\displaystyle x^2-a^2=0}$.
Therefore p must be equal to -3.
Hence the equation is
${\displaystyle x^2-9=0}$
and the roots are 3 and -3. So the desired product is -9.