How do you solve the quadratic #8x^2-4x+5=0# with complex numbers?

1 Answer
Meave60
Dec 16, 2016

#x=(1+3i)/4,##(1-3i)/4#

Explanation:

#8x^2-4x+5=0# is a quadratic equation with the form #ax^2+bx+c#, where #a=8#, #b=-4#, and #c=5#. This equation can be solved using the quadratic formula.

Quadratic Formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the given values.

#x=(-(-4)+-sqrt((-4)^2-4*8*5))/(2*8)#

Simplify.

#x=(4+-sqrt(16-160))/16#

#x=(4+-sqrt(-144))/16#

#x=(4+-12i)/16#

#x=(4+12i)/16,##(4-12i)/16#

Reduce.

#x=(1+3i)/4,##(1-3i)/4#

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