What are the real and complex solutions of the polynomial equation?
x^4 - 41x^2= -400
x^4 - 41x^2+400 =0
x^2= 41+-sqrt(1681-1600))/2
x^2= 21, or 20
x=+- sqrt 21, or +-sqrt20
looks like four real roots.
To find the real and complex solutions of the polynomial equation x^4 - 41x^2 = -400, we can rearrange it to get:
x^4 - 41x^2 + 400 = 0.
Now, let's factorize the equation by using the substitution method. We can substitute x^2 with a new variable, say y:
y = x^2.
This way, our polynomial equation becomes:
y^2 - 41y + 400 = 0.
Now, we can factorize this quadratic equation:
(y - 25)(y - 16) = 0.
Applying the zero-product property, we can set each factor to zero:
y - 25 = 0 or y - 16 = 0.
Solving these equations, we get:
y = 25 or y = 16.
Since we defined y as x^2, we substitute back to solve for x.
For y = 25:
x^2 = 25,
Taking the square root of both sides, we have:
x = ±5.
For y = 16:
x^2 = 16,
Taking the square root of both sides, we have:
x = ±4.
Therefore, the real solutions to the equation are x = ±5 and x = ±4.
Now, let's find the complex solutions if there are any.
The complex numbers are of the form a + bi, where a and b are real numbers, and i is the imaginary unit, defined as the square root of -1.
To find the complex solutions, we substitute the real solutions back into the original equation and solve for the imaginary part.
For x = 5:
x^4 - 41x^2 = -400,
(5)^4 - 41(5)^2 = -400,
625 - 1025 = -400,
-400 = -400.
This means that x = 5 is a solution, but it does not contribute to any complex solutions.
Similarly, for x = -5:
(-5)^4 - 41(-5)^2 = -400,
625 - 1025 = -400,
-400 = -400.
This means that x = -5 is a solution, but it also does not contribute to any complex solutions.
Hence, the polynomial equation x^4 - 41x^2 = -400 has only real solutions x = ±5 and x = ±4.
No complex solutions are present for this particular equation.
To find the real and complex solutions of the polynomial equation x^4 - 41x^2 = -400, we can rewrite the equation as x^4 - 41x^2 + 400 = 0.
Let's solve this equation step-by-step using factoring:
Step 1: Begin by factoring the left side of the equation. In this case, we can write it as (x^2 - 20)(x^2 - 21) = 0.
Step 2: Set each factor equal to zero and solve for x:
From x^2 - 20 = 0, we can take the square root of both sides to find two solutions: x = ±√20, which simplifies to x = ±2√5.
From x^2 - 21 = 0, we can also take the square root of both sides to find two solutions: x = ±√21.
Therefore, the real solutions of the equation are { -2√5, 2√5, -√21, √21 }.
Step 3: To find the complex solutions, we can rearrange the equation and solve for x using the quadratic formula:
x^2 - 41x + 400 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values a = 1, b = -41, and c = 400 into the quadratic formula, we get:
x = (41 ± √((-41)^2 - 4(1)(400))) / 2(1)
x = (41 ± √(1681 - 1600)) / 2
x = (41 ± √81) / 2
x = (41 ± 9) / 2
This gives us two more possible real solutions:
x = (41 + 9) / 2 = 25
x = (41 - 9) / 2 = 16
Therefore, the additional real solutions are { 25, 16 }.
In conclusion, the real solutions of the equation x^4 - 41x^2 = -400 are { -2√5, 2√5, -√21, √21, 25, 16 }. There are no complex solutions.