Answer:
See attached
Explanation:
In general, equations with even powers of the variable will have two real roots. Here, the exception is x² = 0, which has only x = 0 as a solution. The reason for this is that an even power of a negative number is the same value as that even power of its opposite.
x² = 1 ⇒ x = ±1
x² = 16 ⇒ x = ±4
Equations with odd powers of the variable will have one real root.
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The roots will be rational if the numerical value in the equation is the required power of a rational number. For example,
x³ = 1/27 ⇒ x = 1/3 . . . . a rational root
x³ = 1/26 ⇒ x = 1/∛26 . . . . an irrational root
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Additional comment
The Fundamental Theorem of Algebra tells us the total number of roots in complex numbers is equal to the degree of the polynomial. In this problem, we're concerned with real, rational roots, so the fact that each of these cubic equations has two more irrational complex roots is not relevant.
In the case of x²=0, the two roots are ±0, which is "zero, with multiplicity 2." In our answer here, we have elected to call this one solution, as the two solutions are not distinct.