Question 1169683
solve for y = -2x^2 + 8x - 4
set y = 0 and switch sides to get:
-2x^2 + 8x - 4 = 0
divide both sides of the equation by 2 to get:
-x^2 + 4x - 2 = 0
multiply both sides of the equation by -1 to get:
x^2 - 4x + 2 = 0
factor the equation to get:
x = 3.4142135623731 or x = 0.5857864376269


those values of x are when -2x^2 + 8x - 4 is equal to 0.
check values in outside and in between those zero points.
i tried x = 0, x = 2, x = 4
when x = 0, the result was negative.
when x = 2 the result was positive.
when x = 4, the result was negative.


since you want to know when -2x^3 + 8x - 4 is greater than zero, that would be when .5857864376269 < x < 3.414235623731


the graph of the equation shows this to be true.


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