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Kitty [74]
4 years ago
15

Find all the zeros of the equation

Mathematics
1 answer:
lapo4ka [179]4 years ago
6 0

Divide both sides by -3, and replace x^2 with y. Then

-3x^4+27x^2+1200=0\iff y^2-9y-400=0

Factorize the quadratic in y to get

y^2-9y-400=(y+16)(y-25)=0\implies y=-16,y=25

which in turn means

x^2=-16,x^2=25

But x^2\ge0 for all real x, so we can ignore the first solution. This leaves us with

x^2=25\implies x=\pm\sqrt{25}=\pm5

If we allow for any complex solution, then we can continue with the solution we ignored:

x^2=-16\implies x=\pm\sqrt{-16}=\pm i\sqrt{16}=\pm4i

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