How do you solve for x ? #(x-2)(x-3)=34/33^2#

2 Answers
Jun 24, 2016

#100/33 " and " 65/33#

Explanation:

#(x-2)(x-3) = 34/33^2#
#=> (33x-66)(33x-99) = 34#

being hopeful that #(33x-66) and (33x-99)# are integers , at this point, I note that this equation is basically a factorization of 34 such that #a.b=34# and #a-b = 33#

Naturally, the factors are 34 and 1 or (-1) and (-34).

There are two options :

Case I : #a=34 and b =1\ => x = 100/33#

Case II : #a=-1 and b = -34 \ => x= 65/33#

Jun 25, 2016

#x=65/33 or 100/33#

Explanation:

Let x-3 =a then x-2==a+1

#(x-2)(x-3)=(33+1)/33^2#

#=>(a+1)a=33/33^2+1/33^2#

#=>a^2+a-1/33^2-1/33=0#

#=>(a+1/33)(a-1/33)+1(a-1/33)=0#

#=>(a-1/33)(a+1/33+1)=0#

#=>(a-1/33)(a+34/33)=0#

#a=1/33 and a=-34/33#

when #a=1/33#

then # x-3=1/33#

#x=3+1/33=100/33#

when #a=-34/33#

then # x-3=-34/33#

# x=3-34/33=(99-34)/33=65/33#