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(x+8)/(x2-81)

I reads X+8 over x exponet 2 -81

Thanks 4 the help and I wish I could build my yahoo answers points like you all,lol.

2006-06-20 05:38:34 · 5 answers · asked by Brandon ツ 3 in Science & Mathematics Mathematics

5 answers

Well this is a rational expression, so you have to comsider that it is undefined if the denominator equals 0. So set the denominator equal to zero, solve for x, and we know what numbers can't be used for x.
x^2-81=0 is a difference of two squares
(x+9)(x-9)=0
x+9=0 or x-9=0
x=-9 or x=9
Therefore possible solutions for x are all real numbers except -9,9.

2006-06-20 05:55:46 · answer #1 · answered by Anonymous · 2 1

x= 9
(9+8)/(9^2-81)
=17/(81-81)
=17/0
= infinity


x= -9
(-9+8)/{(-9)^2-81}
= -1/(81-81)
= 17/0
= infinity


x = any irrational number cannot be used either

2006-06-20 13:13:47 · answer #2 · answered by Anonymous · 0 0

(x + 8)/(x^2 - 81)
(x + 8)/((x - 9)(x + 9))

x cannot equal 9 or -9

2006-06-20 16:47:50 · answer #3 · answered by Sherman81 6 · 0 0

9 and -9. Then the right parenthesis will be equal to 0, and division with zero is illegal.

2006-06-20 12:40:50 · answer #4 · answered by User1 2 · 0 0

9 or -9. These would cause division by zero.

2006-06-20 12:40:02 · answer #5 · answered by bequalming 5 · 0 0

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