Answer:
I think it is D. Because x=x and there's infinite possibilities.
8a - 4 = 25 + 6 (a - 8)
(distribute the 6)
8a - 4 = 25 + 6a - 48
(combine like terms 25 - 48)
8a - 4 = 6a - 23
( subtract 6a from both sides)
2a - 4 = -23
(add 4 to both sides)
2a = -19
(divide both sides by 2)
a = - 9.5
////CHECK/////
8 (-9.5) - 4 = 25 + 6 ( -9.5 -8)
(plug each side into a calculator)
-80 = -80
FINAL ANSWER:
a = -9.5
If
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then
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The ODE in terms of these series is
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

We can solve the recurrence exactly by substitution:


So the ODE has solution

which you may recognize as the power series of the exponential function. Then

Step 1: Make equation equal to Zero
3x^2 = 4x + 2
3x^2 - 4x - 2
Step 2: use equation
a= 3 b = -4 c= -2
after plugging in answer should be
(2 + or - sqrt 10) / 3