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

How to find zeros?

User Fecub
by
3.3k points

1 Answer

5 votes

Answer:

x=-5, 5

Explanation:

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

(-3x^4+27x^2+1200=0)/(-3) ==> divide by -3 to remove like terms

x^4-9x^2-400=0

x^4+16x^2-25x^2-400=0

x^2(x^2+16)-25(x^2+16)=0

(x^2-25)(x^2+16)=0

x^2-25=0 x^2+16=0

x^2+5x-5x-25=0 x^2+16-16=0-16

x(x+5)-5(x+5)=0 x^2=-16 ==> the square of a real number can't

(x-5)(x+5)=0 be negative

x-5=0 ==> x=5

x+5=0 ==> x=-5

x=-5, 5

User Pavel Anossov
by
3.7k points