Question #4a68b

Question

833 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-19T07:06:43

#x = 15/4#

Note that as we have #4-x# under a radical, we must have #x<=4# to avoid taking the root of a negative number.

#4+sqrt(10-x) = 6+sqrt(4-x)#

#=> sqrt(10-x) = 2+sqrt(4-x)#

#=> (sqrt(10-x))^2 = (2+sqrt(4-x))^2#

#=> 10-x = 2^2 + 2(2)sqrt(4-x) + (sqrt(4-x))^2#

#=> 10-x = 4 + 4sqrt(4-x) + 4 - x#

#=> 2 = 4sqrt(4-x)#

#=> sqrt(4-x) = 1/2#

#=> (sqrt(4-x))^2 = (1/2)^2#

#=> 4-x = 1/4#

#=> x = 4-1/4#

#:. x = 15/4#

Checking our result:

#4+sqrt(10-15/4) = 4+sqrt(25/4)#

#= 4+5/2#

#= 13/2#

#=6+1/2#

#=6+sqrt(1/4)#

#=6+sqrt(4-15/4)#

as desired