Question

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Find all the zeroes of the equation.

–3x4+ 27x2 + 1200 = 0

Answer 1

Values of x are 4i, -4i, 5 and -5

Explanation:

`3x^4+27x^2 + 1200` We need to find all the zeros (roots) of the above equation.

Let assume that x^4 = u^2 and x^2 = u

Putting values of x^4 and x^2 in the above equation and finding the value of u.

`-3u^2 + 27u+1200=0\\Using \,\,quadratic\,\,equation\,\,to\,\,solve:\\u=(-b\pm√(b^2-4ac))/(2a)\\where\,\, a= -3, b= 27 \,\,and\,\, c=1200\\u=(-(27)\pm√((27)^2-4(-3)(1200)))/(2(-3))\\u=(-27\pm√(729+14,400))/(-6)\\u=(-27\pm√(729+14,400))/(-6)\\u=(-27\pm√(15129))/(-6)\\u=(-27\pm123)/(-6)\\so, \,\, u = (-27+123)/(-6) \,\, and \,\, u  (-27-123)/(-6)\\u= -16 \,\, and \,\, u = 25`

So, values of u are -16 and 25

Putting back the value of u i.e, x^2

x^2 = -16 and x^2 =25

solving

Taking square root on both sides:

√x^2 =√-16 and √x^2 = √25

x = ± 4i (as √-1 =i) and x = ±5

So, values of x are 4i, -4i, 5 and -5.