Question

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Find all the zeroes of the equation.

–3x4+ 27x2 + 1200 = 0

Answer 1

Answer:

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

Step-by-step 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.

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

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.