Question:
In the Binomial expansion of $(a-\frac x2)^6$ the coefficient of $x^3$ is 120 times the coeffiecient of $x^5$. Find the possible values of constant $a$.
I've done so far:
Using the fact that:
$${n \choose r} a^{n-r}\left(\frac{-x}{2}\right)^r$$ Equals to the term where $r$ is the coefficient of $x$ and $n$ is the degree of the binomial
I've reached the following equation:
$$\left(\frac{-20a^3x^3}{960}\right)=\left(\frac{-6a^2x^5}{32}\right)$$
How can I use this to find a value of $a$?
Best Answer
You have solved it correctly. Just compare the coefficients and not the terms including x. $$(\frac{-20a^3}{960})=(\frac{-6a}{32})$$ find a=$\pm$3.