Can the square root of a number be negative in a radical equation?

Question

I solved this radical equation: sqrt(2y+1)+7=4and got that y=4.

When I check the answer by plugging 4 back in I get:

sqrt(2*4+1)+7=4
sqrt(9) = -3
The book says this is invalid and the equation has no solutions, But can't the sqrt(9) be +3 or -3??

Answer 1

It cannot. While it is true that (-3)²=9, √x is defined to be the unique nonnegative real number y such that y²=x. The reason for the restriction to nonnegative values is that, for most purposes, it is considered preferable to have the square root function be an actual function -- that is, to return only one output for each input. If we considered both positive and negative values, then the square root function would return two outputs for each input (rather than one), and would not be a function. Because of this, it is impossible for the square root function to return a negative value, so the equation cited really does have no solutions.

Answer 2

There is a rule that says the radical sqrt cannot take negative values.

ie sqrt(9) = +3 NOT -3.
While 9^(1/2) = -3 OR +3.

Who made that rule, I don't know. But that's the way it is.

Answer 3

A square of a number can be negative, but I don't think the sqrt of a number can be negative.

Therefore, the book is right. The problem have no solution.

Answer 4

The Principal root is the positive real root of a positive number, or the negative real root in the case of odd roots of negative numbers.
√(9) + 7 = 3 + 7 = 10.

Answer 5

sr = square root -3 sr X= -15 sr X = -15 divided by -3 sr X = 5 therefore x = 5 squared x = 25

Answer 6

I believe you are correct, as x^2 = (-x)^2 which implies sqrt(x^2) = sqrt ( (-x)^2) = -x

Sometimes even text books make mistakes.

Answer 7

yes, it can. the square root of any number is always the "range of the absolute values of the given number's factors." According to McDougall Litell, you are right.