Explanation:
so, we have
2x² - 3×sqrt(3x) - 15 = 0
2x² - 15 = 3×sqrt(3x)
now, let's square both sides
(2x² - 15)² = 9×3x = 27x
4x⁴ - 60x² + 225 = 27x
4x⁴ - 60x² - 27x + 225 = 0
that leads to enormous terms and expressions.
now, could it be that the actual problem is
2x² - 3×sqrt(3)×x - 15
hmmm ?
to be considered : with the square root of the variable in the expression this is officially not a polygon ...
because then we could solve for a quadratic equation.
a quadratic equation
ax²c+ bx + c = 0
has the general solutions
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 2
b = -3×sqrt(3)
c = -15
x = (3×sqrt(3) ± sqrt(9×3 - 4×2×-15))/(2×2) =
= (3×sqrt(3) ± sqrt(27 + 120))/4 =
= (3×sqrt(3) ± sqrt(147))/4 =
= (3×sqrt(3) ± sqrt(49×3))/4 =
= (3×sqrt(3) ± 7×sqrt(3))/4
x1 = (3×sqrt(3) + 7×sqrt(3))/4 = 10×sqrt(3)/4 =
= 5/2 × sqrt(3)
x2 = (3×sqrt(3) - 7×sqrt(3))/4 = -4×sqrt(3)/4 =
= - sqrt(3)
about the relationship between zeros and coefficients of the polynomial :
the sum of both zeros = -b/a = 3×sqrt(3)/2
5/2 × sqrt(3) + - sqrt(3) = 5/2 × sqrt(3) - sqrt(3) =
= 3/2 × sqrt(3)
correct.
the product of both zeros = c/a = -15/2
5/2 × sqrt(3) × - sqrt(3) = 5/2 × -3 = -15/2
correct.