Asked by Jon
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)
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)
Answers
Answered by
Reiny
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
This is a straightforward difference of squares setup
(x+25)(x-25) = 0
so x = ± 25
2 real roots, so C
Answered by
Jon
I made it harder than it was thanks. But I just thought that x was imaginary
Answered by
Damon
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.
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