There's only one: -1.31951 (rounded)
The other four are complex.
To find the fifth roots of 4 + 3i, first convert the number to polar form: 4 + 3i = 5∠36.87°. Then, to find the fifth roots, divide the angle by 5: 36.87° / 5 = 7.374°. The fifth roots in polar form are 5∠7.374°, 5∠67.374°, 5∠127.374°, 5∠187.374°, and 5∠247.374°.
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.
What does this question mean? -60 has a real cube root, a real fifth root. In fact a real nth root for all odd n.
The 4th root of 256 is 4
In general, the answer is 4, but only 2 of them are real. For example, the 4th roots of 16 are 2, -2, 2i, and -2i.
There are 3 cube roots of 27. There are 2 square roots of 27 ( or any real number ). There are 4 fourth roots of 27 and so on:)
The discriminant of x2+3x+4 is -7 therfore it has no real roots.
real roots= Overdamped equal roots= critically damped complex roots /imaginary roots = Underdamped
There is no difference between real solutions and real roots.
Each distinct real root is an x-intercept. So the answer is 4.
The real fourth roots are -0.3 and 0.3