Solve the following equation in the complex number system. Please show all of your work.
x^4 – 13x^2 – 90 = 0
To solve the given equation x^4 – 13x^2 – 90 = 0 in the complex number system, we will use a method called substitution.
Let's substitute x^2 with a variable, say y. Then the equation becomes y^2 – 13y – 90 = 0. Now we can solve this quadratic equation for y.
To solve the quadratic equation, we can use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -13, and c = -90. Plugging these values into the quadratic formula, we get:
y = (-(-13) ± √((-13)^2 - 4(1)(-90))) / (2(1))
= (13 ± √(169 + 360)) / 2
= (13 ± √529) / 2
= (13 ± 23) / 2
Therefore, we have two possible values for y:
1. y = (13 + 23) / 2 = 36 / 2 = 18
2. y = (13 - 23) / 2 = -10 / 2 = -5
Since we initially substituted x^2 for y, we can substitute back to find possible values of x:
1. For y = 18:
x^2 = 18
x = √18 or x = -√18
2. For y = -5:
x^2 = -5
x = √(-5) or x = -√(-5)
Now, let's simplify the values of x:
1. For x^2 = 18:
x = √18 or x = -√18
x = √(9 * 2) or x = -√(9 * 2)
x = 3√2 or x = -3√2
2. For x^2 = -5:
x = √(-5) or x = -√(-5)
x = i√5 or x = -i√5
Therefore, the solutions to the equation x^4 – 13x^2 – 90 = 0 in the complex number system are:
x = 3√2, -3√2, i√5, -i√5