Add -7 to both sides so that the equation becomes -3x^2 + 6x + 2 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -6 ± √((6)^2 - 4(-3)(2)) ] / ( 2(-3) )
x = [-6 ± √(36 - (-24) ) ] / ( -6 )
x = [-6 ± √(60) ] / ( -6)
x = [-6 ± 2*sqrt(15) ] / ( -6 )
x = 1 ± -sqrt(15)/3
The answers are 1 + sqrt(15)/3 and 1 - sqrt(15)/3.
The intercepts of the third degree polynomial corresponds to the zeros of the equation
y = d*(x-a)*(x-b)(x-c)
Where a, b and c are the roots of the polynomial and d an adjustment coefficient.
y = d*(x+2)*(x)*(x-3)
Lets assume d = 1, and we get
y = (x+2)*(x)*(x-3) = x^3 - x^2 - 6x
We graph the equation in the attached file.
Well it depends what your asking, considering there’s no question mark.
ab^2x
\<span>ab^2x/ab^2y = --------------- = x/y. the ab^2 factor cancels from numerator
ab^2y and denominator.
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