Binomial Equations
Binomial equations of degree n take the general form $$ ax^n + b = 0 $$ where n is a positive integer and a, b are real numbers with a ≠ 0.
Solving a binomial equation
These equations are relatively easy to solve.
The method consists of isolating the variable and taking the nth root of both sides of the equation.
$$ x^n = - \frac{b}{a} $$
$$ \sqrt[n]{x^n} = \sqrt[n]{ - \frac{b}{a} } $$
$$ x = \sqrt[n]{ - \frac{b}{a} } $$
Two cases must be considered:
- n is odd
If the degree n is odd, the equation has exactly one real solution. - n is even
If the degree n is even, the equation has two real solutions, provided the radicand is non-negative.
Note. If the radicand -b/a is negative and the root index is even, the equation has no real solution, because no real number raised to an even power produces a negative result.
Examples
Example 1
Consider the cubic equation:
$$ 2x^3 + 54 = 0 $$
This equation is of odd degree, so it has one real root.
Isolate the variable:
$$ x^3 = - \frac{54}{2} $$
$$ x^3 = -27 $$
Take the cube root of both sides:
$$ \sqrt[3]{x^3} = \sqrt[3]{ -27 } $$
$$ x = \sqrt[3]{ -27 } $$
The cube root of -27 is -3, since (-3)3 = -27.
$$ x = -3 $$
The equation is now solved.
Example 2
Consider the quartic equation:
$$ 2x^4 + 32 = 0 $$
Isolate x:
$$ x^4 = \frac{-32}{2} $$
$$ x^4 = -16 $$
Take the fourth root of both sides:
$$ \sqrt[4]{x^4} = \sqrt[4]{ -16 } $$
$$ x = \sqrt[4]{ -16 } $$
Here, the root index is even (4) and the radicand is negative.
Therefore, the equation has no real roots.
Example 3
Now consider another quartic equation:
$$ 2x^4 - 32 = 0 $$
Isolate x:
$$ x^4 = \frac{32}{2} $$
$$ x^4 = 16 $$
Take the fourth root of both sides:
$$ \sqrt[4]{x^4} = \sqrt[4]{ 16 } $$
$$ x = \sqrt[4]{ 16 } $$
In this case, the index is even (4) and the radicand is positive (16). The equation therefore has two real roots.
The fourth root of 16 is ±2, since (-2)4 = 16 and 24 = 16.
$$ x = \pm 2 $$
Thus, the binomial equation has two real solutions.
