How do you solve #abs(2x+9)=abs(7x-2)#?

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Answer 1 · 2018-06-26T11:33:01

Solution: #x=11/5, x= -7/9#

#|2 x+ 9| =|7 x- 2| # . Squaring both sides we get,

#(2 x+ 9)^2 =(7 x- 2)^2 # or

#4 x^2+36 x +81 =49 x^2-28 x +4 # or

#45 x^2- 64 x -77= 0 ; a=45 ,b=-64 ,c=-77#

Discriminant # D= b^2-4a c = 64^2-4*77*45=17956#

Quadratic formula: #x= (-b+-sqrtD)/(2a) #

#:. x= (64+-sqrt 17956)/(2*45)= 32/45 +- 134/90 # or

#x = 32/45 +- 67/45:. x = 99/45=11/5 and x= -35/45=-7/9 #

Solution: #x=11/5, x= -7/9# [Ans]