How do you solve # 1/18x^2 -1/9x-1=0# using the quadratic formula?

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Answer 1 · 2017-02-12T06:53:01

#x=1+-sqrt19#

Quadratic formula gives the solution of equation #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#

In #1/18x^2-1/9x-1=0#, we have #a=1/18#, #b=-1/9# and #c=-1#.

Hence, #x=(-(-1/9)+-sqrt((-1/9)^2-4(1/18)(-1)))/(2xx1/18)#

= #(1/9+-sqrt(1/81+2/9))/(1/9)#

= #(1/9+-sqrt(19/81))/(1/9)#

= #(1+-sqrt19)#

Alternatively, we could have simplified #1/18x^2-1/9x-1=0# by multiplying each term by #18# to get

#x^2-2x-18=0# and then solution would have been

#x=(-(-2)+-sqrt((-2)^2-4xx1xx(-18)))/2#

= #(2+-sqrt(4+72))/2=1+-sqrt19#