<s>[INST] 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.
 [/INST] There might be a problem with your dot product.
We have
$$(a+xb)\cdot(a-b) = a^2 -a\cdot b + x \ a \cdot b -x \ b^2=0 \\ \implies a^2 +(x-1)\left(2a^2 \cdot \cos \frac{2\pi}{3}\right) -4xa^2=0$$
$$\implies 1-(x-1)-4x=0\implies x=\frac 25$$
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