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sqrt9x+55=x+5 x=-6,5 ??
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is it?\[\sqrt{9x+55}=x+5\]
yup
square both sides to lose the radical\[9x+55=x^2+10x+25\]set equal to 0\[x^2+x-30=0\]factor\[(x+6)(x-5)=0\]zero-factor theorem\[x=-6 \cup x=5\]you are correct once again. :})
wait, wait wait... we forgot to check. when you square you introduce so-called extraneous solutions. check the -6 it does not work
\[\sqrt{-54+55}=-6+5\]\[1=-1\]this implies that x=-6 is not a solution
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but fo the x=5\[\sqrt{45+55}=5+5\]\[10=10\]so x=5 is the only solution
ahhhhhh ok, i always forget about checking lol
btw, i wrote that you do introduce extraneous... i should have written you MAY introduce. . .
lol, tis ok!
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