Question

3+square root of (x+3)=x

Answer 1

\[3+\sqrt{x+3}=x\]

Answer 2

yes

Answer 3

by squaring both sides we get:
9+x+3=\[x ^{2}\]

Answer 4

\[x ^{2}-x-12=0\]
\[here \sum =-1,product equals=-12\]

Answer 5

so \[x ^{2}-4x+3x-12=0\]
x(x-4)+3(x-4)=0
therefore x-4=0 and x+3=0
so x=4 or x=-3

Answer 6

but answer shows to be {6}

Answer 7

isolate that radical firs before you square both sides... subtract 3 from both sides first.

Answer 8

\[\huge 3+\sqrt{x+3}=x\]
\[\huge \sqrt{x+3}=x-3\]
\[\huge x+3=x^2-6x+9\]
\[\huge 0=x^2-7x+6\]
solve from here...