Finding the Solutions of the Equation x4 + 3x2 – 20 = 0
To solve the equation x4 + 3x2 – 20 = 0, we can make a substitution to simplify the problem. Let us set:
y = x2
This transforms our equation into:
y2 + 3y – 20 = 0
Now, we can solve this quadratic equation using the quadratic formula:
y = (-b ± √(b2 – 4ac)) / 2a
Where:
- a = 1
- b = 3
- c = -20
Plugging in these values:
y = (-3 ± √(32 – 4 * 1 * (-20))) / (2 * 1)
Calculating the discriminant:
32 – 4 * 1 * (-20) = 9 + 80 = 89
Now substituting back into the quadratic formula:
y = (-3 ± √89) / 2
This gives us two possible values for y:
- y1 = (-3 + √89) / 2
- y2 = (-3 – √89) / 2
Next, we need to convert these y values back to x values, recalling our substitution y = x2.
Thus:
y1 = x2 => x = ±√y1
y2 = x2 => x = ±√y2
Calculating these gives us:
- x = ±√((-3 + √89) / 2)
- x = ±√((-3 – √89) / 2)
Since y2 could produce negative values for x2, we will only consider values from y1. Hence, the solutions to the original equation are approximately:
- x ≈ ± 3.3
In conclusion, the solutions for the equation x4 + 3x2 – 20 = 0 are:
- x ≈ 3.3
- x ≈ -3.3