Solve the given problem that involves binomial theorem

[chwala]

Homework Statement

See attached

Relevant Equations

Binomial theorem

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...

 

[pasmith]

The binomial expansion of $(1 + x)^\alpha$ is only valid for $|x| < 1$. $$(4 + 3x)^{1.5} = 4^{1.5}\left(1 + \frac{3x}{4}\right)^{1.5}.$$

 

[chwala]

The binomial expansion of $(1 + x)^\alpha$ is only valid for $|x| < 1$. $$(4 + 3x)^{1.5} = 4^{1.5}\left(1 + \frac{3x}{4}\right)^{1.5}.$$

I should have expressed my answer as,

$|x|<\dfrac{4}{3}$