Just to clarify... the two expressions are equivalent for restricted y-values. What does that mean, and why are they equivalent?
[-1+sqrt(3)]1/4
There are no 'real' numbers that can do that. The numbers are 3 + j sqrt(3) and 3 - j sqrt(3). ( ' j ' is the square root of negative 1 )
sqrt(28) + sqrt(63) =sqrt(7 x 4) + sqrt(7 x 9) =[ sqrt(7) x sqrt(4) ] + [ sqrt(7) x sqrt(9) ] =2 sqrt(7) + 3 sqrt(7) =5 sqrt(7) = 13.229 (rounded)
4i square root of 16 is 4, but negative square roots aren't real so the i means its imaginary.
sqrt 6400 = 80 so negative sqrt = -80
sqrt(-5) + sqrt(-20) = sqrt(-5) + sqrt(4)*sqrt(-5)= sqrt(-5) + 2*sqrt(-5) = 3*sqrt(-5)3*sqrt(5)*i or 6.7082*iwhere i is the imaginary square root of -1.
1/(5+sqrt(6)) is equal to about .134237
It is sqrt(3)*[1 + sqrt(2)]/3 or [1 + sqrt(2)]/sqrt(3)
sqrt(a)+sqrt(b) is different from sqrt(a+b) unless a=0 and/or b=0. *sqrt=square root of
-24 = 4 x 6 x -1 Sqrt -24 = sqrt 4 x sqrt 6 x sqrt -1 = 2 x sqrt 6 x i = 2i root6
[-1+sqrt(3)]1/4
There are no 'real' numbers that can do that. The numbers are 3 + j sqrt(3) and 3 - j sqrt(3). ( ' j ' is the square root of negative 1 )
sqrt(28) + sqrt(63) =sqrt(7 x 4) + sqrt(7 x 9) =[ sqrt(7) x sqrt(4) ] + [ sqrt(7) x sqrt(9) ] =2 sqrt(7) + 3 sqrt(7) =5 sqrt(7) = 13.229 (rounded)
121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12
4i square root of 16 is 4, but negative square roots aren't real so the i means its imaginary.
Since the numbers are complex, let's re-write the question as:i*sqrt(4) * i*sqrt(9)i^2, by the good graces of algebra, is equal to -1, so your answer simplifies to:= -1 * sqrt(4) * sqrt(9)= -1 * 2 * 3= -6
sqrt 6400 = 80 so negative sqrt = -80