How do you solve #sqrt(20-X) + 8= sqrt( 9-X) +11#?

2 Answers
May 19, 2016

Answer:

#x=80/9#

Explanation:

Given,

#sqrt(20-x)+8=sqrt(9-x)+11#

Subtract #8# from both sides.

#sqrt(20-x)+8color(white)(i)color(red)(-8)=sqrt(9-x)+11color(white)(i)color(red)(-8)#

#sqrt(20-x)=sqrt(9-x)+3#

Square both sides to get rid of the radical signs.

#(sqrt(20-x))^2=(sqrt(9-x)+3)^2#

#(sqrt(20-x))(sqrt(20-x))=(sqrt(9-x)+3)(sqrt(9-x)+3)#

Simplify.

#20-x=(9-x)+6sqrt(9-x)+9#

#20-x=9-x+6sqrt(9-x)+9#

#2=6sqrt(9-x)#

Solve for #x#.

#1/3=sqrt(9-x)#

#(1/3)^2=(sqrt(9-x))^2#

#1/9=9-x#

#x=color(green)(|bar(ul(color(white)(a/a)color(black)(80/9)color(white)(a/a)|)))#

May 19, 2016

Answer:

#x = 80/9#

Explanation:

Regrouping and powering
#(sqrt[20 - x] - sqrt[9 - x])^2 =(11 - 8)^2#
giving
#29 - 2 sqrt[9 - x] sqrt[20 - x] - 2 x = 9#
Regrouping and powering
#(2 sqrt(9 - x) sqrt(20 - x))^2 =(29 - 9 - 2 x)^2#
giving
#9x=80#
Solving for #x# we get #x = 80/9#