point of Intersection

No
ans = 0

eliminate y from (1) & substitute in (2)

u vl get an eqn in 4th degree....

now sum of roots is = -(coeff of x³)/(coeff of x⁴)

from (i) y = x²-x-4
putting this in (ii)

x⁴ + x² + 16 - 2x³ + 8x - 8x² = x²-x-4 - 15x + 36

=> x⁴ - 2x³ - 8x² + 24x - 16 = 0

we need to find the sum of REAL roots (there maybe imaginary roots as well)

the above exp is factorised as

(x-2)²(x²+2x-4) = 0 (both sides give real roots)

So the sum of roots is 2 (from (x-2)²) + (-2) (from(x²+2x-4)

so the sum of roots is zero

so the sum of x-coordinates of pts of interseciton is zero.