Question 633773


Solve
 √(x+1)+5=x+4


{{{sqrt(x + 1) + 5 = x + 4}}}


{{{sqrt(x + 1) = x + 4 - 5}}} --- Subtracting 5


{{{sqrt(x + 1) = x - 1}}}


{{{(sqrt(x + 1))^2 = (x - 1)^2}}} ------ Squaring both sides of equation


{{{x + 1 = x^2 - 2x + 1}}}


{{{x^2 - 2x - x + 1 - 1 = 0}}}


{{{x^2 - 3x = 0}}}


x(x - 3) = 0


x = 0    or   x - 3 = 0


{{{x = 0}}}    or   {{{x = 3}}}


Howewer, when substituting 0 for x in original equation, x = 0 proves to be an ENTRANEOUS solution, so the only solution is {{{highlight_green(x = 3)}}}


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