Take the equation, substitute y = x^2 into it to reduce it to y^2 + 20y -125 = 0
You find y = 5 & -25
So, x^2 = 5 & -25 the 5 yields real roots +-5^(1/2)
but the -25 yields =- (-25)^(1/2) which yields +- 5i (2 complex roots)
Doug J.
asked 04/17/19Find all complex zeros for the following function.
h(x) = x^4+20x²-125
Take the equation, substitute y = x^2 into it to reduce it to y^2 + 20y -125 = 0
You find y = 5 & -25
So, x^2 = 5 & -25 the 5 yields real roots +-5^(1/2)
but the -25 yields =- (-25)^(1/2) which yields +- 5i (2 complex roots)
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