How do you solve #sqrt(x+9)=sqrt3+sqrtx# and check your solution?

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Answer 1 · 2016-10-15T14:45:34

#x= 3#

square both sides:

#(sqrt(x + 9))^2 = (sqrt(3) + sqrt(x))^2#

#x + 9 = 3 + x + 2sqrt(3x)#

#6 = 2sqrt(3x)#

Square again to get rid of the square root on the right.

#6^2 = (2sqrt(3x))^2#

#36 = 4(3x)#

#36 = 12x#

#x =3#

Finally, don't forget to check your solution in the original equation.

#sqrt(9 + 3) =^? sqrt(3) + sqrt(3)#

#sqrt(12) =^? 2sqrt(3)#

#sqrt(4 xx 3) =^? 2sqrt(3)#

#2sqrt(3) = 2sqrt(3)#

Hopefully this helps!