How do you solve #12x^2 - 7x + 1 = 0 #?

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Answer 1 · 2016-03-31T17:35:43

The solutions are:

#color(green)(x= 1/3#

#color(green)(x= 1/4#

#12x^2 - 7x + 1 = 0 #

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=12, b=-7, c=1#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (-7)^2-(4* 12 * 1)#

# = 49 - 48 = 1#

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-7)+-sqrt(1))/(2*12) = (7+- 1 )/24#

#x= (7+1) / 24= 8/24 = 1/3#

#x= (7-1) / 24= 6/24 = 1/4#