# What is the equation with roots that are twice the roots of ax^2 + bx + c = 0

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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.`

LetÂ `alpha and beta ` are roots of the equation `ax^2+bx+c=0`

Then by relation between roots and coefficients

`alpha+beta=-b/a`

`alphaxx beta=c/a`

We wish to find an equation whose roots are `2 alpha and 2 beta.`

Thus

`2alpha+2beta=2(alpha+beta)=2(-b/a)=-(2b)/a`

`2alphaxx2beta=4alphaxxbeta=4(c/a)=(4c)/a`

Thus required equation is

`x^2-(-(2b)/a)x+(4c)/a=0`

`ax^2+2bx+4c=0`

ans.