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

2 Answers
Noah G
Apr 6, 2017

#{4}#

Explanation:

We start by squaring both sides.

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

#x + 5 = 25 - 10sqrt(x) + x#

#10sqrt(x) = 20#

#sqrt(x) = 20/10#

#sqrt(x) = 2#

We square both sides one more time.

#x = 4#

Now we check our answer.

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

#3 = 5 - 2 color(green)(√)#

Practice exercises

#1#. Solve for #x#. Make sure to verify your solutions because they may or not be extraneous.

a) #sqrt(2x - 2) = sqrt(x) + 1#
b) #sqrt(3x - 5) = sqrt(x + 6) + 1#

Solutions

#1#. a) #x = 9#
b) #x = 10#

Hopefully this helps, and good luck!

Gió
Apr 6, 2017

I got #x=4#

Explanation:

We can try write it as:
#sqrt(x+5)+sqrt(x)=5#
square both sides:
#(sqrt(x+5)+sqrt(x))^2=5^2#
#x+5+2sqrt(x+5)sqrt(x)+x=25#
#2sqrt(x+5)sqrt(x)=20-2x#
square again:
#4x(x+5)=400-80x+4x^2#
#cancel(4x^2)+20x=400-80x+cancel(4x^2)#
#100x=400#
#x=400/100=4#

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