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solve using the square root property

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Greg, are you sure the equation is x2+216=0, with addition? If that's the right equation, then subtracting 216 from both sides yields x2= -216. This equation has no real solutions, because you can't square a real number and get a negative number. But it does have two complex solutions, namely plus or minus the square root of -216, which is plus or minus i*(the square root of 216).

If, however, the equation was supposed to be x2 -216=0, with subtraction instead of addition, then we can add 216 to both sides to get x2=216, so that x would be plus or minus the square root of 216, as you said. If you're confused about what the square root of 216 is, you can either (a) use a calculator to approximate it to any desired number of decimal places, or (b) simplify the square root of 216 by looking for factors that are perfect squares. A moment's thought shows you that 216=36*6, so the square root of 216 is the same as (square root of 36)*(square root of 6), but the square root of 36 is 6, so we just get 6*(square root of 6). If you don't immediately recognize that 216 is divisible by 36, you can try looking for smaller perfect squares like 4 and 9 first; you'll just have to simplify in more than one step.

Does this answer whatever question you had?


B.S. Theoretical Mathematics, MIT



Well Greg, thanks for asking.

x^2 + 216 = 0

Step by step solving we get:
x^2 + 216 = 0
squareroot(x^2 + 216) = squareroot (0)
sqrt(x^2) + sqrt(216) = 0
x + sqrt(216) = 0
x = -sqrt(216)

I hope this helps :)


Timothy, this isn't correct. In your second line, you took the square root of both sides of the equation, which is fine, but then in your third line you made the classic mistake of assuming squareroot(a+b)=squareroot(a)+squareroot(b) But this is definitely false! You can't break up square roots over sums or differences. To see why, just look at a simple example; choose a=8 and b=8. squareroot(8+8)=squareroot(16)=4 but squareroot(8)+squareroot(8) =2squareroot(8) definitely isn't 4! (In fact, it's greater than 4.) So be careful! You can only break up square roots over products and quotients, not over sums and differences. To solve the given equation, you first must pull 216 to the other side, and THEN take square roots.

Thanks for correcting me. I appreciate it. I was doubtful in the first place but it's good that I know now. You're right though. I was reading your answer because I thought the same thing, the problem must have been written incorrectly.