If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to 
Let an angle between the vectors $a$ and $b$ be $\frac{2\pi}{3}$. If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to $\frac23/\frac25/\frac13/\frac15$?

My attempt: $b^2=4a^2$. Also, $$a^2+\frac{|a||b|}{2}-\frac{x|a||b|}{2}-b^2=0$$$$\implies a^2+\frac{|a|*2|a|}{2}-\frac{x|a|*2|a|}{2}-4a^2=0$$$$\implies 1+1-x-4=0$$$$\implies x=-2$$ But it doesn't match with the options.
 A: Remember, $a$ and $b$ are vectors, not scalars, so $a^2 = 4b^2$ doesn't make a lot of sense. Much of your working suffers from the same issue, where it is unclear what is a vector, what is a scalar, and hence what can even be added together.
Our information is:


*

*The angle between the two vectors is $\frac{2\pi}{3}$,

*$|b| = 2|a|$,

*$a + xb \perp a - b$.


All of these pieces of information are relevant, including the first piece, which you seem to be neglecting. From 1, I get
$$a \cdot b = |a| |b| \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}|a||b|.$$
Using 2, we get
$$a \cdot b = -|a|^2.$$
Using 3,
\begin{align*}
0 &= (a + xb) \cdot (a - b) \\
&= |a|^2 + (x - 1)(a \cdot b) - x|b|^2 \\
&= |a|^2 - (x - 1)|a|^2 - 4x|a|^2 \\
&= |a|^2(2 - 5x).
\end{align*}
Note $a \neq 0$, as we need an angle to be defined. We must therefore have $x = \frac{2}{5}$.
A: There might be a problem with your dot product.
We have
$$(a+xb)\cdot(a-b) = a^2 -a\cdot b + x \ a \cdot b -x \ b^2=0 \\ \implies a^2 +(x-1)\left(2a^2 \cdot \cos \frac{2\pi}{3}\right) -4xa^2=0$$
$$\implies 1-(x-1)-4x=0\implies x=\frac 25$$
