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dexar [7]
3 years ago
6

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

Mathematics
1 answer:
guajiro [1.7K]3 years ago
5 0

Answer:

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

Step-by-step 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

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