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x4 + x2 = 1
x4 + x2 - 1 = 0 (a quartic equation: 4 roots).

Let x2 = t

t2 + t - 1 = 0 (use the quadratic formula)
t = [-1 ± √[12 - 4(1)(-1)]]/2(1) = [-1 ± √(1 + 4)]/2 = (-1 ± √5)/2
t = (-1 + √5)/2 ≈ 0.62 or
t = (-1 - √5)/2 ≈ -1.62

Since x = ± √t we have:
x = ± √0.62 = ± 0.79
x = ± √-1.62 = ± i√1.62 = ± 1.27i


Or you can find real roots of the equation by graphing y = x4 + x2 - 1 (in a standard window), and find real zeros of the function, which are ± 0.79. But you have to work more in order two find the other two imaginary roots such as:

Let the other roots be a + bi and a - bi.

Since there is not a third power term, the sum of the roots is 0. So that

0.79 + -0.79 + a + bi + a - bi = 0
2a = 0
a = 0 (the imaginary roots are bi and -bi)

The product of the roots equals to -1/1 = -1 (the constant/the coefficient of x4). So that

(0.79(-0.79)(bi)(-bi) = -1
-b2i2 = -1/-0.6241 (substitute -1 for i2)
b2 = 1.61
b = ± √1.61 = ± 1.27 .
the imaginary roots are ± 1.27i.

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15y ago

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