How do you solve #sqrt(x-7) = 7 - sqrt(x)#?

1 Answer
Apr 3, 2016

Answer:

#x=16#

Explanation:

#color(blue)(sqrt(x-7)=7-sqrtx#

Take the square of both sides to remove the radical sign in the l.h.s

#rarr(sqrt(x-7))^2=(7-sqrtx)^2#

#rarrx-7=(7-sqrtx)^2#

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

#rarrx-7=7^2-2(7)(sqrtx)+sqrtx^2#

#rarrx-7=49-14sqrtx+x#

Subtract #x# both sides

#rarrcancelx-7cancel(-x)=49-14sqrtx+cancel(x-x#

#rarr-7=49-14sqrtx#

Subtract #49# both sides

#rarr-7-49=cancel49-14sqrtxcancel(-49#

#rarr-56=-14sqrtx#

Divide both sides by #-14#

#rarr(-56)/-14=(cancel(-14)sqrtx)/cancel(-14#

#rarr4=sqrtx#

Square both sides

#rarr4^2=sqrtx^2#

#color(green)(rarr16=x#