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

2 Answers
A. S. Adikesavan · Veddesh #phi#
Apr 7, 2016

x = 4.

Explanation:

I have deleted my inclusion of x=0 as a solution. Thanks to Veddesh, for duly pointing out the mistake..
#x>=0#.
#(sqrt(5+x))^2=(5-sqrtx)^2#
This simplifies to #sqrtx=2#. So, x = 4.

Veddesh #phi#
Apr 7, 2016

#x=4#

Explanation:

#color(blue)(sqrt(5+x)+sqrtx=5#

Subtract #sqrtx# both sides

#rarrsqrt(5+x)+cancel(sqrtx-sqrtx)=5-sqrtx#

#rarrsqrt(5+x)=5-sqrtx#

Square both sides to get rid of the radical sign

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

In the right hand side use the formula

#color(brown)((a-b)^2=a^2-2ab+b^2#

So,

#rarr5+x=25-10sqrtx+x#

Subtract #x# and #5# both sides

#rarrcancel(5+x-5-x)=25-10sqrtx+x-5-x#

#rarr0=20-10sqrtx#

Rewrite the equation in standard form

#rarr20-10sqrtx=0#

Subtract #20# both sides

#rarr20-10sqrtx-20=-20#

#rarr-10sqrtx=-20#

Divide both sides by #-10#

#rarr(cancel(-10)sqrtx)/cancel(-10)=(-20)/-10#

#rarrsqrtx=2#

Square both sides

#rarrx=2^2#

#color(green)(rArrx=4#

Related questions
See all questions in Radical Equations
Impact of this question
12794 views around the world
You can reuse this answer
Creative Commons License