How do you solve #sqrt(20-X) + 8= sqrt( 9-X) +11#?
2 Answers
Explanation:
Given,
#sqrt(20-x)+8=sqrt(9-x)+11#
Subtract
#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
#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)|)))#
Explanation:
Regrouping and powering
giving
Regrouping and powering
giving
Solving for