If both roots of the equation  ${\mathrm{x}}^2-2\mathrm{ax}+{\mathrm{a}}^2-1=0$ lie in the interval (-3, 4)then sum of the integral parts of ‘a’ is

${\text{x}}^{\text{2}}{\text{-2ax\hspace{0.17em}+\hspace{0.17em}a}}^{\text{2}}\text{-1\hspace{0.17em}=\hspace{0.17em}0\hspace{0.17em}\hspace{0.17em}}\Rightarrow {\left(\text{x-a}\right)}^{\text{2}}\text{=1}$
Roots are a – 1 and a + 1
Both roots lie in (-3, 4)
$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a-1\hspace{0.17em}>\hspace{0.17em}-3,\hspace{0.17em}\hspace{0.17em}a+1\hspace{0.17em}<\hspace{0.17em}4\hspace{0.17em}\hspace{0.17em}ie}$
$\text{-2\hspace{0.17em}\hspace{0.17em}<\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}<\hspace{0.17em}3\hspace{0.17em}\hspace{0.17em}}\Rightarrow \left[\text{a}\right]=-2,-1,0,1,2$ Whose sum is zero.