Asked by Robin
x^2-4x+32=0.
would the final answer be 2 +or- 2 sqrt7 ?
would the final answer be 2 +or- 2 sqrt7 ?
Answers
Answered by
Lena
x^2-4x+32=0
a =1
b= -4
c= 32
4 +or- sq root (-4)^2 -4(1)(32)
4 +or- sq root 16-128
4 +or- sq root -112
you can't take the sq root of a (-) number so you can't solve this :/
How did you get 2 +or- 2 sqrt7.
Perhaps I went wrong somewhere.
a =1
b= -4
c= 32
4 +or- sq root (-4)^2 -4(1)(32)
4 +or- sq root 16-128
4 +or- sq root -112
you can't take the sq root of a (-) number so you can't solve this :/
How did you get 2 +or- 2 sqrt7.
Perhaps I went wrong somewhere.
Answered by
Steve
hmmm. what's b^2-4ac? 16-128 = -112
So, x = 2 ± 2√(-7) = 2 ± 2√7 i
gotta watch those pesky minus signs
So, x = 2 ± 2√(-7) = 2 ± 2√7 i
gotta watch those pesky minus signs
Answered by
Robin
the original question was
x^2-4x=32
can you help me solve?
x^2-4x=32
can you help me solve?
Answered by
Steve
as I suspected, that makes it x^2-4x-32=0
since 32=8*4,
(x-8)(x+4) is the factorization
since 32=8*4,
(x-8)(x+4) is the factorization
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