How do you solve #- d = 3 sqrt (d - 2)#?

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Answer 1 · 2016-03-07T04:06:22

You must first isolate the square root.

#-d = 3sqrt(d - 2)#

#-d/3 = sqrt(d - 2)#

You must square both sides of the equation to get rid of the equation.

#(-d/3)^2 = (sqrt(d - 2))^2#

#d^2/9 = d - 2#

#d^2 = 9(d - 2)#

#d^2 = 9d - 18#

#d^2 - 9d + 18 = 0#

#(d - 6)(d - 3) = 0#

#d = 6 and d= 3#

Check your solutions back in the equation. Neither work, so there is no solution. Therefore, the solution is #{O/}#.

Practice exercises:

Solve for x:

a) #sqrt(2x + 1) = sqrt(4x) - 1#

b). #sqrt(2x + 2) + sqrt(9x - 2) = 12#

Challenge problem

Solve the following system.

#sqrt(8x + 1) + sqrt(y + 4) = 3x - 1#
#sqrt(11x + 3) - sqrt(y - 1) = x + 1#

Good luck!