How do you solve #sqrt(2x+5) - sqrt(x-2) = 3#?

1 Answer
Jan 14, 2016

#x=2# and #x=38#

Explanation:

Square both sides

#(sqrt(2x+5)-sqrt(x-2))(sqrt(2x+5)-sqrt(x-2))=(3)^2#

Expanding the left side we will have

#(2x+5)-2sqrt(x-2)sqrt(2x+5)+(x-2)=9#

Rewrite the product of the radicals on the left hand side

#2x+5-2sqrt((x-2)(2x+5))+x-2=9#

Combine like terms and expand under the radical on the left

#3x+3-2sqrt(2x^2+x-10)=9#

Subtract #9# from both sides and add #2sqrt(2x^2+x-10)# to both sides

#3x-6=2sqrt(2x^2+x-10)#

Now square both sides

#(3x-6)(3x-6)=(2sqrt(2x^2+x-10))^2#

Expand

#9x^2-36x+36=4(2x^2+x-10)#

Distribute the #4# on the right side

#9x^2-36x+36=8x^2+4x-40#

Subtract #8x^2# from both sides

Subtract #4x# from both sides

Add #40# to both sides

#x^2-40x+76=0#

This will factor to

#(x-2)(x-38)=0#

So #x-2=0#

#x=2#

or #x-38=0#

#x=38#