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Find all the zeros of the equation

-3x^4+27^2+1200=0
if you could show yourworkthat would be great :3

User Philnext
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1 Answer

3 votes

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√(25)=\pm5

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


x^2=-16\implies x=\pm√(-16)=\pm i√(16)=\pm4i

User Cduck
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