How do you solve #4-sqrt(x-3)=sqrt(x+5)#?

Question

2324 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-10T18:36:35

#4-sqrt(x-3) = sqrt(x+5)#

Square both sides giving
#16 - 8sqrt(x-3) + (x-3) = x+5#

#13 - 8sqrt(x-3) + cancel(x) = cancel(x) + 5#

#-8sqrt(x-3) = -8#

#sqrt(x-3) = 1#

Square both sides again
#x-3 = 1#

#x = 4#

Verify by substituting #x=4# back into the original equation.

Answer 2 · 2015-04-10T19:07:14

#x=4#

#4-sqrt(x-3) = sqrt(x+5)#

Squaring both sides

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

As #(a-b)^2 = a^2-2ab+b^2#

So,

#4^2-8sqrt(x-3)+[sqrt(x-3)]^2 = [sqrt(x+5)]^2#

#16-8sqrt(x-3)+(x-3) = x+5#

#16+x-3-x-5-8sqrt(x-3)=0 #

#16-8-8sqrt(x-3) =0 #

#8-8sqrt(x-3)=0 #

#8[1-sqrt(x-3)] =0 #

#1-sqrt(x-3)=0 #

#sqrt(x-3) =1 #

Again Squaring both sides

#[sqrt(x-3)]^2 =1 ^2#

#x-3 =1 #

#x =1+3 #

#x=4#