How do you solve #12c^2+8c+1=0#?

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Answer 1 · 2016-05-04T21:16:32

#c = -1/6# or #c = -1/2#

Find a pair of factors of #12# with sum #8#.

The pair #6, 2# works.

Hence:

#0 = 12c^2+8c+1 = (6c+1)(2c+1)#

which has zeros:

#c = -1/6# and #c = -1/2#

Footnote

Notice that if we reverse the order of the coefficients of the terms, then we get:

#c^2+8c+12 = (c+6)(c+2)#

which has zeros #-6# and #-2#, the reciprocals of the original polynomial.

This is no accident.

For example, the zeros of the polynomial:

#2x^3 - 5x^2 + x + 4#

are the reciprocals of the zeros of:

#4x^3 + x^2 - 5x + 2#