The square root operation is not a function because for each value of y there can be 2 values of x - the principal square root and its negative. This can only be rectified by limiting the range of the opearation to the principal or positive square root.
Furthermore, it also depends on the domain of the function. If y<4 then the square root is not defined within Real numbers.
So, for y ≥ 4, x = +sqrt(y-4) is a function.
sqrt(100x4y4) = sqrt(100)*sqrt(x4)*sqrt(y4) = 10*x2*y2
Assuming the domain and range are suitably defined, then yes. If not, then no.
x=y4 /2
√(40 x12y9) = 6.3246√(y)* x6 *y4 or 6.3246x6y4.5 ■
y4
The GCF is y4
You cannot solve it since only one side of an equation is given. If the equation was y4 + 5y2 - 84 = 0 then y4 + 12y2 - 7y2 - 84 = 0 or y2*(y2 + 12) - 7*(y2 + 12) = 0 or (y2 - 7)*(y2 + 12) = 0 then y2 = 7 or y2 = - 12 y = +or- sqrt(7) and, if you are in the complex domain, also y = +or- i*sqrt(12) where i is the imaginary square root of -1.
4x-y4 what = 0
The simplest equation would be y4 = 0
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
(x2 + y2)(x + y)(x - y) = x4 - y4.
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