A student claims that the equation the !-x = 3 (square root of) has no solution, since the square root of a negative number does not exist. Why is this argument wrong? (The teacher does not want us to work the problem out, only tell why it is wrong.)
If you mean
sqrt(-x) = 3
then as long as x is negative, you are taking the square root of a positive number.
The student's argument is incorrect because it is based on a common misconception about the square root of a negative number. While it is true that the square root of a negative number is not a real number, it does exist as a complex number.
To explain why the argument is wrong, we can refer to the concept of complex numbers. In the complex number system, we have a special imaginary unit denoted as "i," where i^2 = -1. With this imaginary unit, we can define the square root of a negative number.
In the given equation, -x = 3√(-1), we can rewrite it as -x = 3i. Dividing both sides by -1, we get x = -3i.
So, the solution to the equation is x = -3i, which does exist as a complex number. This means that the equation is solvable, contrary to the student's claim.
Hence, the student's argument is incorrect because it overlooks the existence of complex numbers and falsely assumes that the square root of a negative number does not exist.