Asked by Daequan
The equation (24x2+25x−47/ax−2)=−8x−3−(53/ax−2) is true for all values of x≠(2/a), where a is a constant.
What is the value of a?
A) -16
B) -3
C) 3
D) 16
What is the value of a?
A) -16
B) -3
C) 3
D) 16
Answers
Answered by
Daequan
Need help Mrs.Sue, 5th grade math
Answered by
MediaBuff
Holy toledo! This is fifth grade math?! Yikes!
Answered by
Damon
Do you mean?
(24x^2+25x−47) /(ax−2)=−8x−3−[53/(ax−2)]
if so
24x^2+25x−47 = (ax−2)(-8x-3) -53
24x^2+25x−47 = -8ax^2 +16x -3ax +6 - 53
32 x^2 + (9+3a)x = 0
x(32 x +9 +3a) = 0
3 a = - 9 - 32x
I have an error or problem has a typo
I bet it is supposed to be 3 a = -9
or a =-3
(24x^2+25x−47) /(ax−2)=−8x−3−[53/(ax−2)]
if so
24x^2+25x−47 = (ax−2)(-8x-3) -53
24x^2+25x−47 = -8ax^2 +16x -3ax +6 - 53
32 x^2 + (9+3a)x = 0
x(32 x +9 +3a) = 0
3 a = - 9 - 32x
I have an error or problem has a typo
I bet it is supposed to be 3 a = -9
or a =-3
Answered by
Damon
(24x^2+25x−47) /(ax−2)=−8x−3−[53/(ax−2)]
if it were
(24x^2+25x−47) /(ax−2)=+8x−3−[53/(ax−2)]
then the x^2 term would go away later and the a=-3 would be perfect.
typo I am sure
if it were
(24x^2+25x−47) /(ax−2)=+8x−3−[53/(ax−2)]
then the x^2 term would go away later and the a=-3 would be perfect.
typo I am sure
Answered by
Damon
OH
24x^2+25x−47 = -8ax^2 +16x -3ax +6 - 53
(24 +8a) x^2 + (9+3a)x = 0
a =-3 makes both = 0
24x^2+25x−47 = -8ax^2 +16x -3ax +6 - 53
(24 +8a) x^2 + (9+3a)x = 0
a =-3 makes both = 0
Answered by
Damon
No, not grade 5 :)
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