Can you please find the zeros for
F(x)=2x^4+17x^2+8?
1 answer:
2x⁴+17x²+8
Let X = x² →→(x = + or -√X )
The quadratic becomes : 2.X² + 17.X + 8. Now solve for X:
X' = [-b+√(b²-4ac)]/2a and X" = [-b-√(b²-4ac)]/2a
Plug I the values of a, b and c and you will find;
X' = -1/2 and X" = - 8
Since X = √x → →X' = √( -1/2) [imaginary roots]
and X" = √-8 →→ X" = 2√-2 [imaginary roots]
Then:
1st ROOT: X' = - √1/2.i
2nd ROOT: X" = + √1/2.i
3rd ROOT: X" = + 2√2.i
4th ROOT: X" = - 2√2.i
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