How many real sixth roots does 64 have? 0, or 1, or 2
The Correct Answer and Explanation is:
The number 64 has 2 real sixth roots. Here’s an explanation:
To find the sixth roots of a number, we want to solve for xxx in the equation x6=64x^6 = 64×6=64. We can express 64 as 64=2664 = 2^664=26, so the equation becomes x6=26x^6 = 2^6×6=26. This gives us the principal root and the other possible roots from the complex plane.
Step-by-step solution:
- Expressing 64 in polar form:
64 is a positive real number, so we can express it in polar form as 64=64(cos0∘+isin0∘)64 = 64(\cos 0^\circ + i\sin 0^\circ)64=64(cos0∘+isin0∘), or simply 64=64ei0∘64 = 64e^{i0^\circ}64=64ei0∘. - Using De Moivre’s Theorem:
The sixth roots of a complex number are given by the formula: xk=r1/6(cos(θ+360∘k6)+isin(θ+360∘k6))x_k = r^{1/6} \left( \cos \left(\frac{\theta + 360^\circ k}{6}\right) + i \sin \left(\frac{\theta + 360^\circ k}{6}\right) \right)xk=r1/6(cos(6θ+360∘k)+isin(6θ+360∘k)) where rrr is the modulus (in this case, 646464), θ\thetaθ is the argument (for 646464, it’s 0), and kkk takes integer values from 0 to 5, representing the different roots. - Calculating the roots:
The modulus of the root is 641/6=264^{1/6} = 2641/6=2, and the argument of the roots is 360∘k6=60∘k\frac{360^\circ k}{6} = 60^\circ k6360∘k=60∘k, where k=0,1,2,3,4,5k = 0, 1, 2, 3, 4, 5k=0,1,2,3,4,5.- For k=0k = 0k=0, the root is 2(cos0∘+isin0∘)=22(\cos 0^\circ + i \sin 0^\circ) = 22(cos0∘+isin0∘)=2 (which is the principal real root).
- For k=3k = 3k=3, the root is 2(cos180∘+isin180∘)=−22(\cos 180^\circ + i \sin 180^\circ) = -22(cos180∘+isin180∘)=−2 (which is the other real root).
Thus, the two real sixth roots of 64 are 222 and −2-2−2.
Conclusion:
There are exactly 2 real sixth roots of 64: 222 and −2-2−2.
