2 and -2.
The 4th root of 256 is 4
The real ones are -5 and +5.
If "a" is positive, it will have two fourth roots, one will be positive and one will be negative it will have one fifth root, which will be positive. If "a" is negative, it will have one fourth root, which will be negative. it will have one fifth root, which will be negative.
the answer is 256:]
256.
The 4th root of 256 is 4
There are no real square roots of -256. But using complex numbers the square roots of -256 are 16i and -16i.
The real fourth roots are -0.3 and 0.3
No, it is not.
The real fourth roots of 16 are the values of ( x ) that satisfy the equation ( x^4 = 16 ). The solutions can be found by taking the fourth root of both sides, resulting in ( x = \pm 2 ). Therefore, the real fourth roots of 16 are ( 2 ) and ( -2 ).
The answer will depend on the form of the fourth root. Positive real numbers will have two fourth roots which are real and two that are complex. Complex numbers will have four complex roots. However, none of these can be "simplified" in the normal sense of the term.
5 and -5
The real ones are -5 and +5.
If "a" is positive, it will have two fourth roots, one will be positive and one will be negative it will have one fifth root, which will be positive. If "a" is negative, it will have one fourth root, which will be negative. it will have one fifth root, which will be negative.
Upto 4. If the coefficients are all real, then it can have only 0, 2 or 4 real roots.
256 has two square roots: 16 and -16.
4 or -4 to the fourth power equal 256