How do you solve #sqrt (100 - d^2) = 10 - d#?

2 Answers

We have #10-d>=0# or #10>=d#

Hence

#sqrt(100-d^2)=10-d#

#sqrt((10+d)(10-d))=sqrt((10-d)^2)#

For #d!=10# we have that

#sqrt((10+d)/(10-d))=1#

Take squares in both sides

#(10+d)/(10-d)=1#

#10+d=10-d#

#d=0#

Hence the solutions are #d=0# and #d=10#

Jun 1, 2018

Answer:

#d = 0# and #10#

Explanation:

#sqrt(100-d^2) = 10 - d#

First, square both sides:
#(sqrt(100-d^2))^2 = (10 - d)^2#

#100 - d^2 = 100 - 20d + d^2#

Subtract #color(blue)100# from both sides of the equation:
#100 - d^2 quadcolor(blue)(-quad100) = 100 - 20d + d^2 quadcolor(blue)(-quad100)#

#-d^2 = -20d + d^2#

Add #color(blue)(d^2)# to both sides of the equation:
#-d^2 quadcolor(blue)(+quadd^2) = -20d + d^2 quadcolor(blue)(+quadd^2)#

#0 = 2d^2 -20d#

Factor out a #color(blue)(2d)#:
#0 = 2d(d-10)#

#2d = 0# and #d - 10 = 0#

#d = 0# and #d = 10#

#-------------------#

Now plug in both solutions to make sure they are really solutions:

First plug in #0#:
#sqrt(100-d^2) = 10 - d#

#sqrt(100-0) = 10 - 0#

#sqrt(100) = 10#

#10 = 10#

This is true. Therefore, #0# is a solution.

Now plug in #10#:
#sqrt(100-d^2) = 10 - d#

#sqrt(100-10^2) = 10 - 10#

#sqrt(100-100)=0#

#sqrt0=0#

#0=0#

This is also a solution.

Hope this helps!