How do you solve #\sqrt { x } + 1= \sqrt { 5x + 1}#?

2 Answers
Douglas K.
Aug 27, 2017

Given: #sqrtx + 1= sqrt(5x + 1)#

Square both sides:

#x + 2sqrt(x)+ 1 = 5x+1#

Combine like terms:

#-4x+2sqrtx = 0#

Divide both sides by -2:

#2x-sqrtx = 0#

We can write x as #(sqrtx)^2#:

#2(sqrtx)^2-sqrtx= 0#

Factor:

#sqrtx(2sqrtx-1) = 0#

This separates into two equations:

#sqrtx = 0 and 2sqrtx-1 = 0#

Only 1 equation needs simplification:

#sqrtx = 0 and sqrtx = 1/2#

Square both:

#x = 0 and x = 1/4#

Check:

#sqrt0 + 1= sqrt(5(0) + 1)# and #sqrt(1/4) + 1= sqrt(5(1/4) + 1)#

#1 = 1# and #3/2 = 3/2#

Both roots check.

George C.
Aug 27, 2017

#x=0" "# or #" "x = 1/4#

Explanation:

Given:

#sqrt(x)+1 = sqrt(5x+1)#

Square both sides of the equation, noting that this may introduce extraneous solutions, to get:

#x+2sqrt(x)+1 = 5x+1#

Subtract #x+1# from both sides to get:

#2sqrt(x) = 4x#

Note that #x=0# is a solution. For other solutions, divide both sides by #4sqrt(x)# to get:

#1/2 = sqrt(x)#

Then square both sides and transpose to get:

#x = 1/4#

Check both of our derived solutions:

#sqrt(color(blue)(0))+1 = 0+1 = 1 = sqrt(1) = sqrt(0+1) = sqrt(5(color(blue)(0))+1)#

#sqrt(color(blue)(1/4))+1 = 1/2+1 = 3/2 = sqrt(9/4) = sqrt(5/4+1) = sqrt(5(color(blue)(1/4))+1)#

So both are solutions of the original equation.

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