How do you solve #sqrt(2x-2) - sqrtx + 3 = 4#?

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Answer 1 · 2016-02-10T21:42:07

#x=9#

First thing, determine the dominion:

#2x-2>0 and x>=0#

#x>=1 and x>=0#

#x>=1#

The standard way is to put one root in each side of the equality and calculate the squares:

#sqrt(2x-2)-sqrt(x)+3=4#

#sqrt(2x-2)=1+sqrt(x)#,

squaring:

#(sqrt(2x-2))^2=(1+sqrt(x))^2#

#2x-2=1+2sqrt(x) +x#

Now, you have only one root. Isolate it and square it again:

#x-3=2sqrt(x)#,
We must remember that #2sqrt(x)>=0# then #x-3>=0# also.

This means the dominion has changed to #x>=3#

squaring:

#x^2-6x+9=4x#

#x^2-10x+9=0#

#x=(10+-sqrt(10^2-4*9))/2#

#x=(10+-sqrt(64))/2#

#x=(10+-8)/2#

#x=5+-4#

#x=9 or x=1#,

Only the solution #x=9# is valid.