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What are the real and complex solutions of the polynomial equation?
x^4–41x^2=–400
To solve the polynomial equation, we can rearrange it to the form:
x^4 - 41x^2 + 400 = 0
Let's substitute y = x^2 to simplify the equation:
y^2 - 41y + 400 = 0
Now we have a quadratic equation in y. We can solve it by factoring or by using the quadratic formula. Since the quadratic doesn't factor easily, let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -41, and c = 400.
y = (-(-41) ± √((-41)^2 - 4(1)(400))) / (2(1))
y = (41 ± √(1681 - 1600)) / 2
y = (41 ± √81) / 2
y = (41 ± 9) / 2
We have two possible solutions for y:
y1 = (41 + 9) / 2 = 25
y2 = (41 - 9) / 2 = 16
Now, let's substitute the values of y back into the equation y = x^2:
For y1 = 25:
25 = x^2
Taking square root of both sides, we get:
x1 = ±√25
x1 = ±5
For y2 = 16:
16 = x^2
Taking square root of both sides, we get:
x2 = ±√16
x2 = ±4
Therefore, the real solutions to the polynomial equation x^4–41x^2=–400 are x = -5, -4, 4, and 5.