How do you find all the real and complex roots of #f(x) = 4x^4 + 4x^2 + 9x – 2#?

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Answer 1 · 2017-01-23T23:33:45

Here's a sketch of how you could solve this algebraically...

#0 = 4x^4+4x^2+9x-2#

#color(white)(0) = (2x^2-ax+b)(2x^2+ax+c)#

#color(white)(0) = 4x^4+(2(b+c)-a^2)x^2+(b-c)ax+bc#

Equating coefficients and rearranging a little, we find:

#{ (b+c = a^2/2+2), (b-c = 9/a), (bc = -2) :}#

Then:

#(a^2/2+2)^2 = (b+c)^2 = (b-c)^2+4bc = 81/a^2-8#

Multiplied out:

#(a^2)^2/4+2a^2+4 = 81/a^2-8#

Hence (multiplying through by #4a^2#) and rearranging:

#(a^2)^3+8(a^2)^2+48(a^2)-324 = 0#

Use Cardano's method to find Real root of this cubic:

#a^2 = 1/3(-8+root(3)(5590+90sqrt(3921))+root(3)(5590-90sqrt(3921))) ~~ 3.6060#

We can choose the positive square root:

#a = sqrt(1/3(-8+root(3)(5590+90sqrt(3921))+root(3)(5590-90sqrt(3921))))#

Then:

#b = 1/2(a^2/2+2+9/a)#

#c = 1/2(a^2/2+2-9/a)#

Leaving two quadratics to solve:

#2x^2-ax+1/2(a^2/2+2+9/a) = 0#

#2x^2+ax+1/2(a^2/2+2-9/a) = 0#