How do you solve #2( x - 1) + 3\sqrt { x - 1} - 2= 0#?

1 Answer
Y
May 9, 2017

Note that the given equation is quadratic-like, factor and solve.

Explanation:

First we can see that the given equation is quadratic like, the degree of the variable term is two times the degree of the second term:
#(x-1)=(x-1)^1# and #sqrt(x-1)=(x-1)^(1/2)#

We can treat #sqrt(x-1)# as a variable, so let #y=sqrt(x-1)#:
#2y^2+3y-2=0#
which we can simply factor:
#(2y-1)(y+2)=0#
so we get that:
#2y-1=0# or #y+2=0#
which give us that #y = 1/2# or #y=-2#

By substituting these values of #y# into #y=sqrt(x-1)#:
#1/2=sqrt(x-1)#
#1/4=x-1#
#5/4=x#
and
#-2=sqrt(x-1)#
which we can identify as not possible because the square root of a number cannot be negative.

Therefore our only solution is #x=5/4#

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