Solve the given problem that involves binomial theorem

AI Thread Summary
The discussion focuses on solving a problem using the binomial theorem, specifically expanding (4 + 3x)^{1.5}. The expansion yields terms such as 8, 9x, and \dfrac{27}{16} x^2, with the condition that x ≠ -\dfrac{4}{3}. Part (c) involves multiplying this expansion by (1 + ax)^2 to derive a quadratic equation for a. The resulting equation simplifies to 8a^2 + 18a - 5 = 0, leading to solutions a_1 = 0.25 and a_2 = -2.5. The validity of the binomial expansion is noted to require |x| < \dfrac{4}{3}.
chwala
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Homework Statement
See attached
Relevant Equations
Binomial theorem
1686664447486.png


part (a)

##(4+3x)^{1.5} = 2^3+ 9x+ \left[\dfrac {1}{2} ⋅ \dfrac {3}{2} ⋅\dfrac {1}{2}⋅\dfrac {1}{2}⋅9x^2\right]+ ...##

##(4+3x)^{1.5}=8+9x+\dfrac {27}{16} x^2+...##part (b)

##x≠-\dfrac {4}{3}##part (c)

##(8+9x+\dfrac {27}{16} x^2+...)(1+ax)^2 = \dfrac{107}{16} x^2##

...

##8a^2+18a+\dfrac {27}{16}=\dfrac{107}{16}##

##8a^2+18a=\dfrac{80}{16}##

##128a^2+288a-80=0##

##8a^2+18a-5=0##

##a_1=0.25##

##a_2= -2.5##

Bingo!

Any other way welcome guys...
 
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The binomial expansion of (1 + x)^\alpha is only valid for |x| &lt; 1. <br /> (4 + 3x)^{1.5} = 4^{1.5}\left(1 + \frac{3x}{4}\right)^{1.5}.
 
pasmith said:
The binomial expansion of (1 + x)^\alpha is only valid for |x| &lt; 1. <br /> (4 + 3x)^{1.5} = 4^{1.5}\left(1 + \frac{3x}{4}\right)^{1.5}.
I should have expressed my answer as,

##|x|<\dfrac{4}{3}##
 
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