1. 2x^3-12x = 0

x(2x^2-12) = 0

2x^2-12=0 (since x!= 0)

x^2 = 6 (after squaring both side)

x = ±sqrt(6) is the solution.

4x^2 + 5 = 0

4x^2 = -5 (subtracting 5 from both sides)

x^2 = -1.25 (dividing both side by 4)

x = sqrt(-1.25) = 0.587519346 or x = -sqrt(-1.25) = -0.587519346 is the solution.

8y^2 + 5y - 4 = 0

8y^2 + 5y - 4 = 0 (changing the expression to a quadratic equation)

y(8y + 5) = 4(4) (multiplying both side with 4 on the rhs)

y = -1/4 (dividing both side by 8)

-3y^3 + 2y + 1 = 0

-3y^3 + 2y + 1 = 0 (solving as cubic equation)

y = (-b ± sqrt(b^2 - 4ac))/2a (using the formula)

b^2 - 4ac = -8 < 0 (since b^2 is always positive and 4ac is always positive)

Thus, the solution does not exists. (The formula does not apply when the discriminant [b^2 - 4ac] is less than 0)

Since there is no solution for the equation, we would write the answer as "no solution".

I hope this helps!

Answer:

hmm the answer is

A.

Step-by-step explanation:

Sheesh