ill do anything for an answer solve the equation for p. identify any extraneous solutions sqrt p=-1
\[\sqrt{p} = -1\]
If this is what I think it is, there are not any real solutions. To undo a square root, you need to square it. When you do that to the one side you have to do it to the other side. When you square the -1, you get a positive 1. But if you fill a positive 1 back into your equation to check it, it doesn't work out, because the square root of 1 is 1, not -1. I'm very confused here.
what does extraneous mean?
p = square root of -1 which has been designated by mathematicians as "i".
im not sure if that helps
Here's a definition of extraneous solution from MathWords: http://www.mathwords.com/e/extraneous_solution.htm A solution of a simplified version of an equation that does not satisfy the original equation. Watch out for extraneous solutions when solving equations with a variable in the denominator of a rational expression, with a variable in the argument of a logarithm, or a variable as the radicand in an nth root when n is an even number.
How could p equal the square root of -1? The square root is applied to the p. You have to square the square root to solve for p. You can't just take the square root of the -1 without also taking the square root of the square root of p\[\sqrt{\sqrt{p}}=\sqrt{-1}\]
so its just no solution?
here are the choices a-1 is the solution the the original problem.1 is an extranous solution b1 is the solution the the original problem. c1 is the solution the the original problem.-1 is an extranous solution d no solution
I would have to say that there is no solution. I just can't see it working out any other way.
ok thank you
Are those answers quoted exactly? Because a, b, and c have "the the" in the answers.
no
I know that an extraneous solution is reached by squaring both sides p = 1 That is extraneous because if we then take square roots we get sq root (p) = 1 which is NOT the original equation.
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