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Find all the zeros of the equation
-3x^4 + 27x^2 + 1200 = 0

User EuRBamarth
by
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1 Answer

6 votes

Answer:


\displaystyle x=-5,\ x=5,\ x=4i,\ x=-4i

Explanation:

Biquadratic Equation

It's a fourth-degree equation where the terms of degree 1 and 3 are missing. It can be solved for the variable squared as if it was a second-degree equation, and then take the square root of the results

Our equation is


\displaystyle -3x^4+27x^2+1200=0

If we call
y=x^2, our equation becomes a second-degree equation


\displaystyle -3y^2+27y+1200=0

Dividing by -3


\displaystyle y^2-9y-400=0

Factoring


\displaystyle (y-25)(y+16)=0

It leads to these solutions


\displaystyle y=25\ ,\ y=-16

Taking back the change of variable, we have for the first solution


\displaystyle x^2=25\Rightarrow x=-5,x=5

Now for the second solution, we get imaginary (complex) values


\displaystyle x^2=-16\Rightarrow x=4i,\ x=-4i

Summarizing, the four solutions for x are


\displaystyle x=-5,\ x=5,\ x=4i,\ x=-4i

User Tiago Farias
by
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