Question from the Classroom: 10
A student claims that the equation Ö(-x) = 3 has no solution, since the square root of a negative number does not exist. Why is this argument wrong?
Because you could write the answer in imaginary numbers. So technically it does exist. Not sure if that's the exact answer you're looking for, but....
If you're thinking along the terms of a+bi
then the square root of -3 would be
1.73205...i
the dots are just the extra numbers. I assume you'd round it. But keep the curved i at the end.
If you equation is
√-x = 3 then
x = -9 is a real number solution to the equation
isn't it an imaginary number
To explain why the student's argument is wrong, we need to understand the concept of complex numbers. In mathematics, the square root of a negative number is not defined in the set of real numbers, but it does exist in a broader number system called the complex numbers.
Complex numbers are composed of two parts: a real part and an imaginary part. They are written in the form a + bi, where "a" is the real part, "b" is the imaginary part, and "i" is the imaginary unit, which is defined as the square root of -1.
In this case, we have the equation Ö(-x) = 3. To solve this equation, we square both sides to eliminate the square root: (-x) = 3^2 = 9. Multiplying both sides by -1 gives us x = -9.
So, the solution to the equation is x = -9. Although the intermediate step involves the square root of a negative number, we can still find a solution by using the concept of complex numbers. Thus, the argument that the equation has no solution due to the square root of a negative number not existing is incorrect.