What are the real and complex solutions of the polynomial equation?
x^4 - 41x^2 = -400
x^4 - 41x^2 = -400
x^4 - 41x^2 + 400 = 0
(x^2-16)(x^2-25) = 0
x = ±4,±5
steve is correct!!!1//1!!
Well, this equation seems to be quite a handful! But fear not, for I am here to solve it with a touch of humor.
Now, let's see how we can clown around with these solutions!
If we rewrite the equation as x^4 - 41x^2 + 400 = 0, we can start the show. We notice that this equation resembles a quadratic in disguise, looking like (x^2 - 20x + 20)(x^2 + 20x - 20) = 0.
So, we're left with two quadratic equations to solve:
1) x^2 - 20x + 20 = 0
2) x^2 + 20x - 20 = 0
Now, let's put on our colorful wigs and juggle these solutions!
Taking the quadratic formula and employing our silly antics, we find that the solutions to the first equation are:
x = (20 ± √20i - 4) / 2
x = 10 ± √(20i - 4)
For the second equation, these are the solutions we bring to the stage:
x = (-20 ± √(20^2 - 4*(-20)) / 2
x = -10 ± √(440) / 2
And when we simplify these solutions, we get:
For the first equation: x ≈ 10 ± 4.472i
For the second equation: x ≈ -10 ± 14.798
And there you have it, a hilarious performance of solving the real and complex solutions of this polynomial equation. I hope you enjoyed the show!
To find the real and complex solutions of the polynomial equation x^4 - 41x^2 = -400, we can use factoring and the quadratic formula.
Step 1: Rewrite the equation in standard form.
x^4 - 41x^2 + 400 = 0
Step 2: Let's substitute y = x^2 to rewrite the equation in terms of y.
y^2 - 41y + 400 = 0
Step 3: Factor the quadratic equation (y^2 - 41y + 400).
(y - 16)(y - 25) = 0
Step 4: Set each factor equal to zero and solve for y.
y - 16 = 0 or y - 25 = 0
Solving for y:
For y - 16 = 0, we get y = 16.
For y - 25 = 0, we get y = 25.
Step 5: Substitute back x^2 for y.
For y = 16, we get x^2 = 16. Taking the square root of both sides, we have x = ±4.
For y = 25, we get x^2 = 25. Taking the square root of both sides, we have x = ±5.
Therefore, the real solutions to the equation are x = 4 and x = -4, and the complex solutions are x = 5i and x = -5i.
To find the real and complex solutions of the polynomial equation x^4 - 41x^2 = -400, we can solve it by factoring.
1. Start by moving all the terms to one side of the equation:
x^4 - 41x^2 + 400 = 0
2. Notice that this equation is a quadratic in terms of x^2. We can treat x^2 as a variable and solve for it by factoring:
(x^2 - 25)(x^2 - 16) = 0
3. Now we have two separate equations to solve:
x^2 - 25 = 0
x^2 - 16 = 0
4. Solve each equation separately:
For x^2 - 25 = 0:
x^2 = 25
x = ±√25
x = ±5
For x^2 - 16 = 0:
x^2 = 16
x = ±√16
x = ±4
5. Combine all the solutions to get the real solutions:
x = 5, x = -5, x = 4, x = -4
Therefore, the real solutions to the polynomial equation x^4 - 41x^2 = -400 are x = 5, x = -5, x = 4, and x = -4.