If |a|> |a+b+c| prove that there is complex root such

livinglife100s8x

livinglife100s8x

Answered question

2022-04-20

If |a|>|a+b+c| prove that there is complex root such that |z|<2.
Let ax2+bx+c=0 be a quadratic equation where |a|>|a+b+c|, a, b, cR. Prove that this equation has at least one solution zC such that |z|<2.

Answer & Explanation

despescarwh9

despescarwh9

Beginner2022-04-21Added 15 answers

If both roots s,t are real then we have
a(xs)(xt)=ax2a(s+t)x+=0. Now |a|>|aa(s+t)+|1>|1st+st|=|(1s)(1t)|.
Clearly |s| and |t| cannot both be greater than 2. The double root case is straightforward.
If both roots s=u+iv and t=uiv are complex we have
a(x(u+iv))(x(uiv))=ax22aux+a(u2+v2)=0.
Now |a|>|a2au+a(u2+v2)|1>|12u+(u2+v2)|.
If |s| (and hence |t|) are greater than 2 then u2+v2>4.

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