Which describes the number and type of roots of the equation x^2-625=0?
A)1 real root, 1 imaginary root
B)2 real roots, 2 imaginary roots
C)2 real roots
D)4 real roots
I went with A
sqrt x^2= x (which would be the imaginary) and sqrt 625= 25 (the real)
How did you get that answer???
This is a straightforward difference of squares setup
(x+25)(x-25) = 0
so x = ± 25
2 real roots, so C
I made it harder than it was thanks. But I just thought that x was imaginary
By the way, if you get one complex or imaginary root, you get two. They come in pairs, like male and female twins. One has +imaginary number, the other has -the same imaginary number. They are called complex conjugates. a + b i and a - b i and a can be zero.
To determine the number and type of roots of the equation x^2 - 625 = 0, we can start by simplifying the equation:
x^2 - 625 = 0
This equation can be factored using the difference of squares formula:
(x - 25)(x + 25) = 0
Now we have two factors: (x - 25) and (x + 25). To find the roots, we set each factor equal to zero:
x - 25 = 0 --> x = 25
x + 25 = 0 --> x = -25
We find that there are two real roots: x = 25 and x = -25. Since both roots are real, they are not imaginary.
Therefore, the correct answer is C) 2 real roots.