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how do I find zeros of a fn?

 

y=3x^2+5x^4-2x^3

 Aug 22, 2016
 #1
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+5

Solve for x:
5 x^4-2 x^3+3 x^2 = 0

 

Factor x^2 from the left hand side:
x^2 (5 x^2-2 x+3) = 0

 

Split into two equations:
x^2 = 0 or 5 x^2-2 x+3 = 0

 

Take the square root of both sides:
x = 0 or 5 x^2-2 x+3 = 0

 

Divide both sides by 5:
x = 0 or x^2-(2 x)/5+3/5 = 0

 

Subtract 3/5 from both sides:
x = 0 or x^2-(2 x)/5 = -3/5

 

Add 1/25 to both sides:
x = 0 or x^2-(2 x)/5+1/25 = -14/25

 

Write the left hand side as a square:
x = 0 or (x-1/5)^2 = -14/25

 

Take the square root of both sides:
x = 0 or x-1/5 = (i sqrt(14))/5 or x-1/5 = 1/5 (-i) sqrt(14)

 

Add 1/5 to both sides:
x = 0 or x = 1/5+(i sqrt(14))/5 or x-1/5 = -(i sqrt(14))/5

 

Add 1/5 to both sides:
Answer: |x = 0    or    x = 1/5+(i sqrt(14))/5    or    x = 1/5-(i sqrt(14))/5

 Aug 22, 2016

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