Another method is to just notice that if `r` is a root of `ax^2+bx+c=0,` then `2r` is a root of `a(x/2)^2+b(x/2)+c=0,` since `a((2r)/2)^2+b((2r)/2)+c=ar^2+br+c=0` .

If we expand `a(x/2)^2+b(x/2)+c` , **we get the answer** `a/4x^2+b/2x+c=0.`

But this isn't the only answer; any nonzero constant multiple of this equation has the same roots, so in particular, we can get aruv's equation by multiplying by 4. `4(a/4x^2+b/2x+c)=ax^2+2bx+4c.`

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