Question 183953This question is from textbook Glencoe McGraw-Hill: Algebra 2
: Hi, I hope I am submitting this problem under the right section, the chapter I'm currently studying has asymptotes so..
Haha, I've tried this problem so many ways! I feel that the mistake is right under my nose, but I still can't seem to find it. the dashes are fractions.
4/z-2 - z+6/z+1 = 1
I go to a private school, but I am currently being homeschooled for a couple weeks. My teacher wanted us to solve this problem by clearing the fractions.
Here's my most "unjumbled" work:
*original problem*
LCD: (z-2)(z+1)
(z-2)(z+1)/ 1 times 4/(z-2) minus (z-2)(z+1)/1 times (z+6)/(z+1) = (z-2)(z+1)/1 times 1.
4z+4-(z-2)(z+6)=(z-2)(z+1)
4z+4-z^2+6z-2z-12= z^2+z-2z-2
4z+4-z^2+4z-12 = z^2-z-2
-z^2+8z-8 = z^2-z-2
eliminate -z^2+8z-8 on the left side, do the same to the right side like...
0 = z^2-z-2
+z^2-8z-8
_______________ equals
0 = 2z^2-9z-10
not factorable, so I used quadratic formula
a=2, b=-9, c=-10
quadratic formula: [-b (+ or -) sq. root of: b^2 - 4ac] divided by 2a
=[9 (+ or -) sq. root of: (-9)^2 - 4(2)(-10)] all divided by 4.
= [ 9 (+ or -) sq. root of: 161] all divided by 4.
my textbook says that the correct answer should be:
[ 1 (+ or -) sq. root of: 145] all divided by 4.
Mmm.. hopefully this is readable. Please help, it's driving me bonkers! Thanks!
This question is from textbook Glencoe McGraw-Hill: Algebra 2
Answer by edjones(8007)   (Show Source):
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