# SOLUTION: how would you solve this? 3^2+square root*(2^3+1+y)=3^0 i started by simplifing 9+square root*(8+1+y)=1 then 9+square root(9+y)=1 then 9+9*square root*(y)=1 then 9*square root*(

Algebra ->  Algebra  -> Equations -> SOLUTION: how would you solve this? 3^2+square root*(2^3+1+y)=3^0 i started by simplifing 9+square root*(8+1+y)=1 then 9+square root(9+y)=1 then 9+9*square root*(y)=1 then 9*square root*(      Log On

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 Click here to see ALL problems on Equations Question 190628: how would you solve this? 3^2+square root*(2^3+1+y)=3^0 i started by simplifing 9+square root*(8+1+y)=1 then 9+square root(9+y)=1 then 9+9*square root*(y)=1 then 9*square root*(y)=(-8) then i got stuck?!?!? could u help me? thanks! Found 2 solutions by jim_thompson5910, solver91311:Answer by jim_thompson5910(28704)   (Show Source): You can put this solution on YOUR website!Unfortunately, there's a mistake your steps. You made a mistake in saying that Start with the given equation. Square 3 to get 9. Cube 2 to get 8. Raise 3 to the zeroth power to get 1. Combine like terms. Subtract 9 from both sides. Combine like terms. Square both sides (to eliminate the square root) Square -8 to get 64 Subtract from both sides. Combine like terms on the right side. So the possible solution is . However, we need to check it. Check: Start with the given equation. Plug in Square 3 to get 9. Cube 2 to get 8. Raise 3 to the zeroth power to get 1. Combine like terms. Take the square root of 64 to get 8 Combine like terms. Since the equation is false, this means that is NOT a solution. So there are no solutions. Answer by solver91311(17076)   (Show Source): You can put this solution on YOUR website! how would you solve this? 3^2+square root*(2^3+1+y)=3^0 i started by simplifing 9+square root*(8+1+y)=1 then 9+square root(9+y)=1 then 9+9*square root*(y)=1 then 9*square root*(y)=(-8) then i got stuck?!?!? Let's see: That is all correct as far as the algebraic manipulations go, but you should be able to see that you have a serious problem. is a positive number by definition, that is, when you see the radical sign you assume the positive square root. The problem is that there is no way to add a positive number, however small, to 9 and have the result be 1 -- or anything less than 9 for that matter. Be that as it may, let's continue. How you got from: to is beyond me. You cannot separate terms under a radical. Think about Pythagoras' Theorem. If what you did was right, then: and there would be no such thing as a triangle. Your next step should have been to add -9 to both sides: And then square both sides: However, if you substitute 55 for y in the original equation, you get: What happened was that when we squared both sides of the equation, we introduced an extraneous root, and that root must be excluded. Since that was the only root, we can then conclude that the solution set to the given equation is the empty set. John