How to solve the equation, #sqrt(5x+6)=x# ?

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Answer 1 · 2018-03-03T13:47:30

#x = 6#

#sqrt(5x + 6) = x#

Notice that #x >= 0# and #5x + 6 >= 0 => x >= 0#

#sqrt(5x + 6)^2 = x^2#

#5x + 6= x^2#

#x^2 - 5x - 6 = 0#

#(x - 6)(x + 1) = 0#

#x = -1, x = 6#

Since #x >=0#, only #x = 6# works in the original equation.