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December 19, 2014

December 19, 2014

Posted by **Robin** on Tuesday, March 12, 2013 at 4:11pm.

would the final answer be 2 +or- 2 sqrt7 ?

- math -
**Lena**, Tuesday, March 12, 2013 at 5:00pmx^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.

- math -
**Steve**, Tuesday, March 12, 2013 at 5:05pmhmmm. what's b^2-4ac? 16-128 = -112

So, x = 2 ± 2√(-7) = 2 ± 2√7 i

gotta watch those pesky minus signs

- math -
**Robin**, Tuesday, March 12, 2013 at 5:20pmthe original question was

x^2-4x=32

can you help me solve?

- math -
**Steve**, Tuesday, March 12, 2013 at 6:17pmas I suspected, that makes it x^2-4x-32=0

since 32=8*4,

(x-8)(x+4) is the factorization

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