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\[\sqrt{x+10}=x-2\]

you first need to get rid of the radical.

so square both sides.

ok

so you should have x+10 = x^2-4x-4 ?

ok

actualy... i typoed.. it should be \[x+10=x^{2}-4x+4\]

ok i was confused trying to work that out

and the only reason i move to the right is cause I dont want to multiply by a negative 1.

\[x ^{2}-5x-6=0\]

you know how to factor or do the quadratic?

i have my text book here, these confuse me so much

so 2 and 3

ok

now since we know we a negative 5.. we know that 6 has to be negative. so (x+1)(x-6)

so (x+1)(x-6)

solve those as separate equations .. x-6=0 and x+1=0

x=6 x=-1

ok

\[\sqrt{-1+10} =-1 - 2\]

\[\sqrt{6+10} = 6-2\]

sqrt9=-3 sqrt16 = 4

so x = 6 is your only answer? does that make sense?

\[\sqrt{9} =3 \neq -1-2 = -3\] \[\sqrt{16}=4 = 6-2=4\]