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Maxim Wagner
Mathematics
3 October, 21:36
B. x4 + 10x2 + 25 = 0
+5
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Guadalupe
3 October, 21:48
0
Solve it like a normal quadratic equation and take the root of the roots
x^4 + 10x^2 + 25 = 0
Δ = b² - 4. a. c
Δ = 10² - 4. 1. 25
Δ = 100 - 4. 1. 25
Δ = 0
1 real root.
In this case, x' = x'':
x = (-b + - √Δ) / 2a
x' = (-10 + √0) / 2.1
x'' = (-10 - √0) / 2.1
x' = - 10 / 2
x'' = - 10 / 2
x' = - 5
x'' = - 5
Now take the root of this roots
√-5 = i√5
The root of this equation is x = ±i√5
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